Continuous image of a locally compact space is not necessarily locally compact












0












$begingroup$



A topological space is said to be locally compact if each point $xin X$ has at least one neighbourhood which is compact.
Prove continuous image of a locally compact space is not necessarily locally compact.




Attempted solution:



I was able to find a counter example I believe:



$g:(X',tau')to(Y',tau'')$ where $tau$ is the discrete topology and $tau_2$ is the indiscrete topology where $Y'$ is infinite, hence it is not compact(there is no finite open set but $emptyset$).



However I am striving to see why the image is not necessarily locally compact.



Admitting $f:(X,tau)to (Y,tau_1)$ is a continuous function. Then if $yin Y$ then $f^{-1}(y)=x$ since $(X,tau)$ is locally compact then $xin U$ such that $U$ is compact. Them image of a compact set is a compact set so $yin f(U)$ so that $f(U)$ is compact. So the image is locally compact.



Question:



1) Since the author states not "necessarily compact". What am I missing?



2) Is my counterexample right?



Thanks in advance!






general-topology proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your definition of compact and locally compact? The indiscrete space is compact (although not Hausdorff compact) just like any space with finite number of open subsets. Note that in the indiscrete space also whole $X$ is open.
    $endgroup$
    – freakish
    Dec 26 '18 at 12:15










  • $begingroup$
    You may want to read this as well: math.stackexchange.com/questions/1287344/…
    $endgroup$
    – freakish
    Dec 26 '18 at 12:17












  • $begingroup$
    @freakish But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23










  • $begingroup$
    @freakish The link you provided the function is assumed to be surjective. Here there is no assumption whatsoever regarding the function but it is continuous.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:25










  • $begingroup$
    @freakish I am using the cover definition for compactness and locally compactness is defined in the beginning of the post.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:26
















0












$begingroup$



A topological space is said to be locally compact if each point $xin X$ has at least one neighbourhood which is compact.
Prove continuous image of a locally compact space is not necessarily locally compact.




Attempted solution:



I was able to find a counter example I believe:



$g:(X',tau')to(Y',tau'')$ where $tau$ is the discrete topology and $tau_2$ is the indiscrete topology where $Y'$ is infinite, hence it is not compact(there is no finite open set but $emptyset$).



However I am striving to see why the image is not necessarily locally compact.



Admitting $f:(X,tau)to (Y,tau_1)$ is a continuous function. Then if $yin Y$ then $f^{-1}(y)=x$ since $(X,tau)$ is locally compact then $xin U$ such that $U$ is compact. Them image of a compact set is a compact set so $yin f(U)$ so that $f(U)$ is compact. So the image is locally compact.



Question:



1) Since the author states not "necessarily compact". What am I missing?



2) Is my counterexample right?



Thanks in advance!






general-topology proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your definition of compact and locally compact? The indiscrete space is compact (although not Hausdorff compact) just like any space with finite number of open subsets. Note that in the indiscrete space also whole $X$ is open.
    $endgroup$
    – freakish
    Dec 26 '18 at 12:15










  • $begingroup$
    You may want to read this as well: math.stackexchange.com/questions/1287344/…
    $endgroup$
    – freakish
    Dec 26 '18 at 12:17












  • $begingroup$
    @freakish But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23










  • $begingroup$
    @freakish The link you provided the function is assumed to be surjective. Here there is no assumption whatsoever regarding the function but it is continuous.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:25










  • $begingroup$
    @freakish I am using the cover definition for compactness and locally compactness is defined in the beginning of the post.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:26














0












0








0





$begingroup$



A topological space is said to be locally compact if each point $xin X$ has at least one neighbourhood which is compact.
Prove continuous image of a locally compact space is not necessarily locally compact.




Attempted solution:



I was able to find a counter example I believe:



$g:(X',tau')to(Y',tau'')$ where $tau$ is the discrete topology and $tau_2$ is the indiscrete topology where $Y'$ is infinite, hence it is not compact(there is no finite open set but $emptyset$).



However I am striving to see why the image is not necessarily locally compact.



Admitting $f:(X,tau)to (Y,tau_1)$ is a continuous function. Then if $yin Y$ then $f^{-1}(y)=x$ since $(X,tau)$ is locally compact then $xin U$ such that $U$ is compact. Them image of a compact set is a compact set so $yin f(U)$ so that $f(U)$ is compact. So the image is locally compact.



Question:



1) Since the author states not "necessarily compact". What am I missing?



2) Is my counterexample right?



Thanks in advance!






general-topology proof-verification







share|cite|improve this question











$endgroup$





A topological space is said to be locally compact if each point $xin X$ has at least one neighbourhood which is compact.
Prove continuous image of a locally compact space is not necessarily locally compact.




Attempted solution:



I was able to find a counter example I believe:



$g:(X',tau')to(Y',tau'')$ where $tau$ is the discrete topology and $tau_2$ is the indiscrete topology where $Y'$ is infinite, hence it is not compact(there is no finite open set but $emptyset$).



However I am striving to see why the image is not necessarily locally compact.



Admitting $f:(X,tau)to (Y,tau_1)$ is a continuous function. Then if $yin Y$ then $f^{-1}(y)=x$ since $(X,tau)$ is locally compact then $xin U$ such that $U$ is compact. Them image of a compact set is a compact set so $yin f(U)$ so that $f(U)$ is compact. So the image is locally compact.



Question:



1) Since the author states not "necessarily compact". What am I missing?



2) Is my counterexample right?



Thanks in advance!






general-topology proof-verification




general-topology proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 12:19









José Carlos Santos

161k22127232




161k22127232










asked Dec 26 '18 at 12:08









Pedro GomesPedro Gomes

1,8262721




1,8262721












  • $begingroup$
    What is your definition of compact and locally compact? The indiscrete space is compact (although not Hausdorff compact) just like any space with finite number of open subsets. Note that in the indiscrete space also whole $X$ is open.
    $endgroup$
    – freakish
    Dec 26 '18 at 12:15










  • $begingroup$
    You may want to read this as well: math.stackexchange.com/questions/1287344/…
    $endgroup$
    – freakish
    Dec 26 '18 at 12:17












  • $begingroup$
    @freakish But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23










  • $begingroup$
    @freakish The link you provided the function is assumed to be surjective. Here there is no assumption whatsoever regarding the function but it is continuous.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:25










  • $begingroup$
    @freakish I am using the cover definition for compactness and locally compactness is defined in the beginning of the post.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:26


















  • $begingroup$
    What is your definition of compact and locally compact? The indiscrete space is compact (although not Hausdorff compact) just like any space with finite number of open subsets. Note that in the indiscrete space also whole $X$ is open.
    $endgroup$
    – freakish
    Dec 26 '18 at 12:15










  • $begingroup$
    You may want to read this as well: math.stackexchange.com/questions/1287344/…
    $endgroup$
    – freakish
    Dec 26 '18 at 12:17












  • $begingroup$
    @freakish But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23










  • $begingroup$
    @freakish The link you provided the function is assumed to be surjective. Here there is no assumption whatsoever regarding the function but it is continuous.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:25










  • $begingroup$
    @freakish I am using the cover definition for compactness and locally compactness is defined in the beginning of the post.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:26
















$begingroup$
What is your definition of compact and locally compact? The indiscrete space is compact (although not Hausdorff compact) just like any space with finite number of open subsets. Note that in the indiscrete space also whole $X$ is open.
$endgroup$
– freakish
Dec 26 '18 at 12:15




$begingroup$
What is your definition of compact and locally compact? The indiscrete space is compact (although not Hausdorff compact) just like any space with finite number of open subsets. Note that in the indiscrete space also whole $X$ is open.
$endgroup$
– freakish
Dec 26 '18 at 12:15












$begingroup$
You may want to read this as well: math.stackexchange.com/questions/1287344/…
$endgroup$
– freakish
Dec 26 '18 at 12:17






$begingroup$
You may want to read this as well: math.stackexchange.com/questions/1287344/…
$endgroup$
– freakish
Dec 26 '18 at 12:17














$begingroup$
@freakish But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:23




$begingroup$
@freakish But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:23












$begingroup$
@freakish The link you provided the function is assumed to be surjective. Here there is no assumption whatsoever regarding the function but it is continuous.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:25




$begingroup$
@freakish The link you provided the function is assumed to be surjective. Here there is no assumption whatsoever regarding the function but it is continuous.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:25












$begingroup$
@freakish I am using the cover definition for compactness and locally compactness is defined in the beginning of the post.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:26




$begingroup$
@freakish I am using the cover definition for compactness and locally compactness is defined in the beginning of the post.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:26










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your counterexample is wrong because the indiscrete topology is locally compact in your definition, but it can be fixed:



If $tau_d$ is the discrete topology on a set $X$ then $f(x)=x$ is continuous
as a map between $(X,tau_d)$ to $(X,tau)$ where $tau$ is any topology on $X$ we like. This observation does hold.
Also, $(X,tau_d)$ is locally compact as every point $x$ has the compact neighbourhood ${x}$ in $tau_d$.



So let $(X,tau)$ be any non-locally compact topology. (Like $mathbb{Q}$ in the standard topology inherited from $mathbb{R}$ (or its order)) and use that special case. The same "example template" can also be used for metrisable spaces, locally connected spaces etc.



So you then have shown that the continuous image of a locally compact space is not necessarily locally compact (you have at least one concrete example where this is not the case). E.g. the continuous image of a compact space is "necessarily compact", because there is a theorem that says so.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is $mathbb{Q}$ not locally compact with the Euclidian distance? It has been hard for me to understand that. Thanks for your answer!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:56










  • $begingroup$
    @PedroGomes did you try searching this site? There are loads of answers that address this question. Using sequences: in any open rational interval there are rational sequences converging (in the reals) to an irrational number.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 13:01










  • $begingroup$
    I have understood the example. However there is still a problem regarding the my attempted proof. It contradicts the example and I do not understand why. I do not know how to reconcile them. What is failing? How should I correct it? thanks in advance!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 18:54










  • $begingroup$
    @PedroGomes there is no contradiction. You happen to give an example where the image is locally compact and I give one where it is not. The question was to show that it was not necessarily (always) locally compact, so my example was the one you need. Yours shows that the image could be locally compact as well.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:59



















0












$begingroup$

Your example is wrong because every subset is compact in indiscrete topology. Also in the proof that follows $f(U)$ is compact but it need not be a neighborhood of $y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23






  • 1




    $begingroup$
    In any indiscrete space there are only two open sets, so any open cover of any subset has a finite subcover. That means any subset is compact.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:26










  • $begingroup$
    Could you give me an example of a non-locally compact space please?
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:29










  • $begingroup$
    $f(U)$ is intended to be the neighbourhood of $y$ not $x$.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:31










  • $begingroup$
    @PedroGomes Sorry for the typo. I meant neighborhood of $y$ when I typed neighborhood of $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:32











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Your counterexample is wrong because the indiscrete topology is locally compact in your definition, but it can be fixed:



If $tau_d$ is the discrete topology on a set $X$ then $f(x)=x$ is continuous
as a map between $(X,tau_d)$ to $(X,tau)$ where $tau$ is any topology on $X$ we like. This observation does hold.
Also, $(X,tau_d)$ is locally compact as every point $x$ has the compact neighbourhood ${x}$ in $tau_d$.



So let $(X,tau)$ be any non-locally compact topology. (Like $mathbb{Q}$ in the standard topology inherited from $mathbb{R}$ (or its order)) and use that special case. The same "example template" can also be used for metrisable spaces, locally connected spaces etc.



So you then have shown that the continuous image of a locally compact space is not necessarily locally compact (you have at least one concrete example where this is not the case). E.g. the continuous image of a compact space is "necessarily compact", because there is a theorem that says so.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is $mathbb{Q}$ not locally compact with the Euclidian distance? It has been hard for me to understand that. Thanks for your answer!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:56










  • $begingroup$
    @PedroGomes did you try searching this site? There are loads of answers that address this question. Using sequences: in any open rational interval there are rational sequences converging (in the reals) to an irrational number.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 13:01










  • $begingroup$
    I have understood the example. However there is still a problem regarding the my attempted proof. It contradicts the example and I do not understand why. I do not know how to reconcile them. What is failing? How should I correct it? thanks in advance!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 18:54










  • $begingroup$
    @PedroGomes there is no contradiction. You happen to give an example where the image is locally compact and I give one where it is not. The question was to show that it was not necessarily (always) locally compact, so my example was the one you need. Yours shows that the image could be locally compact as well.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:59
















1












$begingroup$

Your counterexample is wrong because the indiscrete topology is locally compact in your definition, but it can be fixed:



If $tau_d$ is the discrete topology on a set $X$ then $f(x)=x$ is continuous
as a map between $(X,tau_d)$ to $(X,tau)$ where $tau$ is any topology on $X$ we like. This observation does hold.
Also, $(X,tau_d)$ is locally compact as every point $x$ has the compact neighbourhood ${x}$ in $tau_d$.



So let $(X,tau)$ be any non-locally compact topology. (Like $mathbb{Q}$ in the standard topology inherited from $mathbb{R}$ (or its order)) and use that special case. The same "example template" can also be used for metrisable spaces, locally connected spaces etc.



So you then have shown that the continuous image of a locally compact space is not necessarily locally compact (you have at least one concrete example where this is not the case). E.g. the continuous image of a compact space is "necessarily compact", because there is a theorem that says so.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is $mathbb{Q}$ not locally compact with the Euclidian distance? It has been hard for me to understand that. Thanks for your answer!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:56










  • $begingroup$
    @PedroGomes did you try searching this site? There are loads of answers that address this question. Using sequences: in any open rational interval there are rational sequences converging (in the reals) to an irrational number.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 13:01










  • $begingroup$
    I have understood the example. However there is still a problem regarding the my attempted proof. It contradicts the example and I do not understand why. I do not know how to reconcile them. What is failing? How should I correct it? thanks in advance!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 18:54










  • $begingroup$
    @PedroGomes there is no contradiction. You happen to give an example where the image is locally compact and I give one where it is not. The question was to show that it was not necessarily (always) locally compact, so my example was the one you need. Yours shows that the image could be locally compact as well.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:59














1












1








1





$begingroup$

Your counterexample is wrong because the indiscrete topology is locally compact in your definition, but it can be fixed:



If $tau_d$ is the discrete topology on a set $X$ then $f(x)=x$ is continuous
as a map between $(X,tau_d)$ to $(X,tau)$ where $tau$ is any topology on $X$ we like. This observation does hold.
Also, $(X,tau_d)$ is locally compact as every point $x$ has the compact neighbourhood ${x}$ in $tau_d$.



So let $(X,tau)$ be any non-locally compact topology. (Like $mathbb{Q}$ in the standard topology inherited from $mathbb{R}$ (or its order)) and use that special case. The same "example template" can also be used for metrisable spaces, locally connected spaces etc.



So you then have shown that the continuous image of a locally compact space is not necessarily locally compact (you have at least one concrete example where this is not the case). E.g. the continuous image of a compact space is "necessarily compact", because there is a theorem that says so.






share|cite|improve this answer









$endgroup$



Your counterexample is wrong because the indiscrete topology is locally compact in your definition, but it can be fixed:



If $tau_d$ is the discrete topology on a set $X$ then $f(x)=x$ is continuous
as a map between $(X,tau_d)$ to $(X,tau)$ where $tau$ is any topology on $X$ we like. This observation does hold.
Also, $(X,tau_d)$ is locally compact as every point $x$ has the compact neighbourhood ${x}$ in $tau_d$.



So let $(X,tau)$ be any non-locally compact topology. (Like $mathbb{Q}$ in the standard topology inherited from $mathbb{R}$ (or its order)) and use that special case. The same "example template" can also be used for metrisable spaces, locally connected spaces etc.



So you then have shown that the continuous image of a locally compact space is not necessarily locally compact (you have at least one concrete example where this is not the case). E.g. the continuous image of a compact space is "necessarily compact", because there is a theorem that says so.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 12:46









Henno BrandsmaHenno Brandsma

109k347115




109k347115












  • $begingroup$
    How is $mathbb{Q}$ not locally compact with the Euclidian distance? It has been hard for me to understand that. Thanks for your answer!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:56










  • $begingroup$
    @PedroGomes did you try searching this site? There are loads of answers that address this question. Using sequences: in any open rational interval there are rational sequences converging (in the reals) to an irrational number.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 13:01










  • $begingroup$
    I have understood the example. However there is still a problem regarding the my attempted proof. It contradicts the example and I do not understand why. I do not know how to reconcile them. What is failing? How should I correct it? thanks in advance!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 18:54










  • $begingroup$
    @PedroGomes there is no contradiction. You happen to give an example where the image is locally compact and I give one where it is not. The question was to show that it was not necessarily (always) locally compact, so my example was the one you need. Yours shows that the image could be locally compact as well.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:59


















  • $begingroup$
    How is $mathbb{Q}$ not locally compact with the Euclidian distance? It has been hard for me to understand that. Thanks for your answer!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:56










  • $begingroup$
    @PedroGomes did you try searching this site? There are loads of answers that address this question. Using sequences: in any open rational interval there are rational sequences converging (in the reals) to an irrational number.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 13:01










  • $begingroup$
    I have understood the example. However there is still a problem regarding the my attempted proof. It contradicts the example and I do not understand why. I do not know how to reconcile them. What is failing? How should I correct it? thanks in advance!
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 18:54










  • $begingroup$
    @PedroGomes there is no contradiction. You happen to give an example where the image is locally compact and I give one where it is not. The question was to show that it was not necessarily (always) locally compact, so my example was the one you need. Yours shows that the image could be locally compact as well.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:59
















$begingroup$
How is $mathbb{Q}$ not locally compact with the Euclidian distance? It has been hard for me to understand that. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:56




$begingroup$
How is $mathbb{Q}$ not locally compact with the Euclidian distance? It has been hard for me to understand that. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:56












$begingroup$
@PedroGomes did you try searching this site? There are loads of answers that address this question. Using sequences: in any open rational interval there are rational sequences converging (in the reals) to an irrational number.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 13:01




$begingroup$
@PedroGomes did you try searching this site? There are loads of answers that address this question. Using sequences: in any open rational interval there are rational sequences converging (in the reals) to an irrational number.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 13:01












$begingroup$
I have understood the example. However there is still a problem regarding the my attempted proof. It contradicts the example and I do not understand why. I do not know how to reconcile them. What is failing? How should I correct it? thanks in advance!
$endgroup$
– Pedro Gomes
Dec 26 '18 at 18:54




$begingroup$
I have understood the example. However there is still a problem regarding the my attempted proof. It contradicts the example and I do not understand why. I do not know how to reconcile them. What is failing? How should I correct it? thanks in advance!
$endgroup$
– Pedro Gomes
Dec 26 '18 at 18:54












$begingroup$
@PedroGomes there is no contradiction. You happen to give an example where the image is locally compact and I give one where it is not. The question was to show that it was not necessarily (always) locally compact, so my example was the one you need. Yours shows that the image could be locally compact as well.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 18:59




$begingroup$
@PedroGomes there is no contradiction. You happen to give an example where the image is locally compact and I give one where it is not. The question was to show that it was not necessarily (always) locally compact, so my example was the one you need. Yours shows that the image could be locally compact as well.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 18:59











0












$begingroup$

Your example is wrong because every subset is compact in indiscrete topology. Also in the proof that follows $f(U)$ is compact but it need not be a neighborhood of $y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23






  • 1




    $begingroup$
    In any indiscrete space there are only two open sets, so any open cover of any subset has a finite subcover. That means any subset is compact.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:26










  • $begingroup$
    Could you give me an example of a non-locally compact space please?
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:29










  • $begingroup$
    $f(U)$ is intended to be the neighbourhood of $y$ not $x$.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:31










  • $begingroup$
    @PedroGomes Sorry for the typo. I meant neighborhood of $y$ when I typed neighborhood of $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:32
















0












$begingroup$

Your example is wrong because every subset is compact in indiscrete topology. Also in the proof that follows $f(U)$ is compact but it need not be a neighborhood of $y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23






  • 1




    $begingroup$
    In any indiscrete space there are only two open sets, so any open cover of any subset has a finite subcover. That means any subset is compact.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:26










  • $begingroup$
    Could you give me an example of a non-locally compact space please?
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:29










  • $begingroup$
    $f(U)$ is intended to be the neighbourhood of $y$ not $x$.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:31










  • $begingroup$
    @PedroGomes Sorry for the typo. I meant neighborhood of $y$ when I typed neighborhood of $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:32














0












0








0





$begingroup$

Your example is wrong because every subset is compact in indiscrete topology. Also in the proof that follows $f(U)$ is compact but it need not be a neighborhood of $y$.






share|cite|improve this answer











$endgroup$



Your example is wrong because every subset is compact in indiscrete topology. Also in the proof that follows $f(U)$ is compact but it need not be a neighborhood of $y$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 12:31

























answered Dec 26 '18 at 12:14









Kavi Rama MurthyKavi Rama Murthy

59.8k42161




59.8k42161












  • $begingroup$
    But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23






  • 1




    $begingroup$
    In any indiscrete space there are only two open sets, so any open cover of any subset has a finite subcover. That means any subset is compact.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:26










  • $begingroup$
    Could you give me an example of a non-locally compact space please?
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:29










  • $begingroup$
    $f(U)$ is intended to be the neighbourhood of $y$ not $x$.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:31










  • $begingroup$
    @PedroGomes Sorry for the typo. I meant neighborhood of $y$ when I typed neighborhood of $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:32


















  • $begingroup$
    But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:23






  • 1




    $begingroup$
    In any indiscrete space there are only two open sets, so any open cover of any subset has a finite subcover. That means any subset is compact.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:26










  • $begingroup$
    Could you give me an example of a non-locally compact space please?
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:29










  • $begingroup$
    $f(U)$ is intended to be the neighbourhood of $y$ not $x$.
    $endgroup$
    – Pedro Gomes
    Dec 26 '18 at 12:31










  • $begingroup$
    @PedroGomes Sorry for the typo. I meant neighborhood of $y$ when I typed neighborhood of $x$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 26 '18 at 12:32
















$begingroup$
But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:23




$begingroup$
But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:23




1




1




$begingroup$
In any indiscrete space there are only two open sets, so any open cover of any subset has a finite subcover. That means any subset is compact.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 12:26




$begingroup$
In any indiscrete space there are only two open sets, so any open cover of any subset has a finite subcover. That means any subset is compact.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 12:26












$begingroup$
Could you give me an example of a non-locally compact space please?
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:29




$begingroup$
Could you give me an example of a non-locally compact space please?
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:29












$begingroup$
$f(U)$ is intended to be the neighbourhood of $y$ not $x$.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:31




$begingroup$
$f(U)$ is intended to be the neighbourhood of $y$ not $x$.
$endgroup$
– Pedro Gomes
Dec 26 '18 at 12:31












$begingroup$
@PedroGomes Sorry for the typo. I meant neighborhood of $y$ when I typed neighborhood of $x$.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 12:32




$begingroup$
@PedroGomes Sorry for the typo. I meant neighborhood of $y$ when I typed neighborhood of $x$.
$endgroup$
– Kavi Rama Murthy
Dec 26 '18 at 12:32


















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