A space $X$ is locally connected if and only if every component of every open set of $X$ is open?
It is claimed on the wikipedia page that a space $X$ is locally connected if and only if every component of every open set of $X$ is open without any proof.
What is the proof behind this fact? Am I correct in assuming this in turn implies that a space is locally connect if and only if the open connected subsets are actually a base for topology?
general-topology
asked Dec 13 '11 at 14:09
Jeffrey Yellow
2612
add a comment |
It is claimed on the wikipedia page that a space $X$ is locally connected if and only if every component of every open set of $X$ is open without any proof.
What is the proof behind this fact? Am I correct in assuming this in turn implies that a space is locally connect if and only if the open connected subsets are actually a base for topology?
general-topology
asked Dec 13 '11 at 14:09
Jeffrey Yellow
2612
add a comment |
It is claimed on the wikipedia page that a space $X$ is locally connected if and only if every component of every open set of $X$ is open without any proof.
What is the proof behind this fact? Am I correct in assuming this in turn implies that a space is locally connect if and only if the open connected subsets are actually a base for topology?
general-topology
asked Dec 13 '11 at 14:09
Jeffrey Yellow
2612
It is claimed on the wikipedia page that a space $X$ is locally connected if and only if every component of every open set of $X$ is open without any proof.
What is the proof behind this fact? Am I correct in assuming this in turn implies that a space is locally connect if and only if the open connected subsets are actually a base for topology?
general-topology
general-topology
asked Dec 13 '11 at 14:09
Jeffrey Yellow
2612
asked Dec 13 '11 at 14:09
Jeffrey Yellow
2612
asked Dec 13 '11 at 14:09
Jeffrey Yellow
2612
asked Dec 13 '11 at 14:09
Jeffrey Yellow
2612
asked Dec 13 '11 at 14:09
Jeffrey Yellow
2612
2612
add a comment |
add a comment |
2 Answers
2
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If $X$ is locally connected and $C$ is a connected component of an open subset $U subseteq X$, then every point $c in C$ contains a connected neighborhood which lies in $U$ (because $X$ is locally connected and $U$ is open). But then it has to lie completely in the connected component of $c$, i.e. in $C$. This shows that $C$ is open.
The proof of the converse is very, very similar. Can you write it down for yourself?
edited Dec 7 at 13:09
Arrow
5,07121445
answered Dec 13 '11 at 15:03
Martin Brandenburg
107k13157326
1
I think you mean $C subseteq U$.
– Vandrin
Oct 11 '16 at 15:52
I think I mean $U subseteq X$.
– Martin Brandenburg
Oct 21 '16 at 13:47
Why $c$ is in $U$ necessarily? Indeed, how can one make sure that there is an open set inside $C$ when $C$ is a component of a point, say $x$? Thanks
– Fardad Pouran
Sep 27 '17 at 11:06
@FardadPouran Because the connected component C build a partition of the subset U. Hence, every point of C has to be in U. The second question doesn't make much sense unless you specify the underlying topology. Be aware that the empty set is open under every topology.
– quallenjäger
Oct 15 '17 at 21:43
add a comment |
Suppose $X$ is locally connected, and take an open set $A$ and a component $C$ of $A$. We want to show that $A backslash C$ is closed in $A$. Take a point $x$ in the closure of $A backslash C$, by local connectedness we can find a connected neighborhood $N$ of $x$. If $x in C$ then $N subseteq C$, but $N$ intersects $A backslash C$, which is absurd. So $C$ is open in $A$, hence in $X$.
The converse follows from the obvious fact that every open set is the union of its components.
As for your second question, the statement is true, but it follows directly from the definition.
answered Dec 13 '11 at 15:07
Paolo Capriotti
40426
add a comment |
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2 Answers
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2 Answers
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If $X$ is locally connected and $C$ is a connected component of an open subset $U subseteq X$, then every point $c in C$ contains a connected neighborhood which lies in $U$ (because $X$ is locally connected and $U$ is open). But then it has to lie completely in the connected component of $c$, i.e. in $C$. This shows that $C$ is open.
The proof of the converse is very, very similar. Can you write it down for yourself?
edited Dec 7 at 13:09
Arrow
5,07121445
answered Dec 13 '11 at 15:03
Martin Brandenburg
107k13157326
1
I think you mean $C subseteq U$.
– Vandrin
Oct 11 '16 at 15:52
I think I mean $U subseteq X$.
– Martin Brandenburg
Oct 21 '16 at 13:47
Why $c$ is in $U$ necessarily? Indeed, how can one make sure that there is an open set inside $C$ when $C$ is a component of a point, say $x$? Thanks
– Fardad Pouran
Sep 27 '17 at 11:06
@FardadPouran Because the connected component C build a partition of the subset U. Hence, every point of C has to be in U. The second question doesn't make much sense unless you specify the underlying topology. Be aware that the empty set is open under every topology.
– quallenjäger
Oct 15 '17 at 21:43
add a comment |
If $X$ is locally connected and $C$ is a connected component of an open subset $U subseteq X$, then every point $c in C$ contains a connected neighborhood which lies in $U$ (because $X$ is locally connected and $U$ is open). But then it has to lie completely in the connected component of $c$, i.e. in $C$. This shows that $C$ is open.
The proof of the converse is very, very similar. Can you write it down for yourself?
edited Dec 7 at 13:09
Arrow
5,07121445
answered Dec 13 '11 at 15:03
Martin Brandenburg
107k13157326
1
I think you mean $C subseteq U$.
– Vandrin
Oct 11 '16 at 15:52
I think I mean $U subseteq X$.
– Martin Brandenburg
Oct 21 '16 at 13:47
Why $c$ is in $U$ necessarily? Indeed, how can one make sure that there is an open set inside $C$ when $C$ is a component of a point, say $x$? Thanks
– Fardad Pouran
Sep 27 '17 at 11:06
@FardadPouran Because the connected component C build a partition of the subset U. Hence, every point of C has to be in U. The second question doesn't make much sense unless you specify the underlying topology. Be aware that the empty set is open under every topology.
– quallenjäger
Oct 15 '17 at 21:43
add a comment |
If $X$ is locally connected and $C$ is a connected component of an open subset $U subseteq X$, then every point $c in C$ contains a connected neighborhood which lies in $U$ (because $X$ is locally connected and $U$ is open). But then it has to lie completely in the connected component of $c$, i.e. in $C$. This shows that $C$ is open.
The proof of the converse is very, very similar. Can you write it down for yourself?
edited Dec 7 at 13:09
Arrow
5,07121445
answered Dec 13 '11 at 15:03
Martin Brandenburg
107k13157326
If $X$ is locally connected and $C$ is a connected component of an open subset $U subseteq X$, then every point $c in C$ contains a connected neighborhood which lies in $U$ (because $X$ is locally connected and $U$ is open). But then it has to lie completely in the connected component of $c$, i.e. in $C$. This shows that $C$ is open.
The proof of the converse is very, very similar. Can you write it down for yourself?
edited Dec 7 at 13:09
Arrow
5,07121445
answered Dec 13 '11 at 15:03
Martin Brandenburg
107k13157326
edited Dec 7 at 13:09
Arrow
5,07121445
edited Dec 7 at 13:09
Arrow
5,07121445
edited Dec 7 at 13:09
Arrow
5,07121445
5,07121445
answered Dec 13 '11 at 15:03
Martin Brandenburg
107k13157326
answered Dec 13 '11 at 15:03
Martin Brandenburg
107k13157326
answered Dec 13 '11 at 15:03
Martin Brandenburg
107k13157326
107k13157326
1
I think you mean $C subseteq U$.
– Vandrin
Oct 11 '16 at 15:52
I think I mean $U subseteq X$.
– Martin Brandenburg
Oct 21 '16 at 13:47
Why $c$ is in $U$ necessarily? Indeed, how can one make sure that there is an open set inside $C$ when $C$ is a component of a point, say $x$? Thanks
– Fardad Pouran
Sep 27 '17 at 11:06
@FardadPouran Because the connected component C build a partition of the subset U. Hence, every point of C has to be in U. The second question doesn't make much sense unless you specify the underlying topology. Be aware that the empty set is open under every topology.
– quallenjäger
Oct 15 '17 at 21:43
add a comment |
1
I think you mean $C subseteq U$.
– Vandrin
Oct 11 '16 at 15:52
I think I mean $U subseteq X$.
– Martin Brandenburg
Oct 21 '16 at 13:47
Why $c$ is in $U$ necessarily? Indeed, how can one make sure that there is an open set inside $C$ when $C$ is a component of a point, say $x$? Thanks
– Fardad Pouran
Sep 27 '17 at 11:06
@FardadPouran Because the connected component C build a partition of the subset U. Hence, every point of C has to be in U. The second question doesn't make much sense unless you specify the underlying topology. Be aware that the empty set is open under every topology.
– quallenjäger
Oct 15 '17 at 21:43
1
1
I think you mean $C subseteq U$.
– Vandrin
Oct 11 '16 at 15:52
I think you mean $C subseteq U$.
– Vandrin
Oct 11 '16 at 15:52
I think I mean $U subseteq X$.
– Martin Brandenburg
Oct 21 '16 at 13:47
I think I mean $U subseteq X$.
– Martin Brandenburg
Oct 21 '16 at 13:47
Why $c$ is in $U$ necessarily? Indeed, how can one make sure that there is an open set inside $C$ when $C$ is a component of a point, say $x$? Thanks
– Fardad Pouran
Sep 27 '17 at 11:06
Why $c$ is in $U$ necessarily? Indeed, how can one make sure that there is an open set inside $C$ when $C$ is a component of a point, say $x$? Thanks
– Fardad Pouran
Sep 27 '17 at 11:06
@FardadPouran Because the connected component C build a partition of the subset U. Hence, every point of C has to be in U. The second question doesn't make much sense unless you specify the underlying topology. Be aware that the empty set is open under every topology.
– quallenjäger
Oct 15 '17 at 21:43
@FardadPouran Because the connected component C build a partition of the subset U. Hence, every point of C has to be in U. The second question doesn't make much sense unless you specify the underlying topology. Be aware that the empty set is open under every topology.
– quallenjäger
Oct 15 '17 at 21:43
add a comment |
Suppose $X$ is locally connected, and take an open set $A$ and a component $C$ of $A$. We want to show that $A backslash C$ is closed in $A$. Take a point $x$ in the closure of $A backslash C$, by local connectedness we can find a connected neighborhood $N$ of $x$. If $x in C$ then $N subseteq C$, but $N$ intersects $A backslash C$, which is absurd. So $C$ is open in $A$, hence in $X$.
The converse follows from the obvious fact that every open set is the union of its components.
As for your second question, the statement is true, but it follows directly from the definition.
answered Dec 13 '11 at 15:07
Paolo Capriotti
40426
add a comment |
Suppose $X$ is locally connected, and take an open set $A$ and a component $C$ of $A$. We want to show that $A backslash C$ is closed in $A$. Take a point $x$ in the closure of $A backslash C$, by local connectedness we can find a connected neighborhood $N$ of $x$. If $x in C$ then $N subseteq C$, but $N$ intersects $A backslash C$, which is absurd. So $C$ is open in $A$, hence in $X$.
The converse follows from the obvious fact that every open set is the union of its components.
As for your second question, the statement is true, but it follows directly from the definition.
answered Dec 13 '11 at 15:07
Paolo Capriotti
40426
add a comment |
Suppose $X$ is locally connected, and take an open set $A$ and a component $C$ of $A$. We want to show that $A backslash C$ is closed in $A$. Take a point $x$ in the closure of $A backslash C$, by local connectedness we can find a connected neighborhood $N$ of $x$. If $x in C$ then $N subseteq C$, but $N$ intersects $A backslash C$, which is absurd. So $C$ is open in $A$, hence in $X$.
The converse follows from the obvious fact that every open set is the union of its components.
As for your second question, the statement is true, but it follows directly from the definition.
answered Dec 13 '11 at 15:07
Paolo Capriotti
40426
Suppose $X$ is locally connected, and take an open set $A$ and a component $C$ of $A$. We want to show that $A backslash C$ is closed in $A$. Take a point $x$ in the closure of $A backslash C$, by local connectedness we can find a connected neighborhood $N$ of $x$. If $x in C$ then $N subseteq C$, but $N$ intersects $A backslash C$, which is absurd. So $C$ is open in $A$, hence in $X$.
The converse follows from the obvious fact that every open set is the union of its components.
As for your second question, the statement is true, but it follows directly from the definition.
answered Dec 13 '11 at 15:07
Paolo Capriotti
40426
answered Dec 13 '11 at 15:07
Paolo Capriotti
40426
answered Dec 13 '11 at 15:07
Paolo Capriotti
40426
answered Dec 13 '11 at 15:07
Paolo Capriotti
40426
40426
add a comment |
add a comment |
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