The one point compactification of a countable $KC$ space$X$ is $KC$ if and only if $X$ is compact
$begingroup$
The bellow Theorem is in " Spaces in which compact subsets are close and the lattice of $T_{1}$ topologies on sets" by " T.Alas , G.Wilson"
The one point compactification of a countable $KC$ space$X$ is $KC$ if and only if $X$ is compact.
proof:
sufficiency: an open subspace of a sequential spaces is sequential.
necessity: supposr that $C$ is a compact subspace of the one point compatification $ Y = X cup { infty }$of $X$. If $ infty notin C $ then $C$ is a compact subspace of $X$ , hence closed and so $ Y - C $ is open in $Y$.
if on the other hand $infty in C $, then if $C$ is notclosed in $Y$, $ C cap X$ is not closed in$X$ and hence there is a sequence ${ x_{n} } $in $C cap X $conerging to some $ p notin C $. Since $X$ is $KC$, the compact set$ S = {p} cup { x_{n} : n in omega } $is closed in $X$ and so $ Y- S $ is a neighborhood of $infty $ and so $infty $ is not accumulation point of${ x_{n} } $, implying that $C$ is not compact, a contradiction.
a): why in the sufficiency part, $X$ is open and he said an open subspace of a sequential spaces is sequential?
b): why in the last line of necessity part, we say that $C$ is not compact?
(c): In The one point compatification, $infty notin X $ but in the proof said that "if $infty in C$? Why?
real-analysis general-topology convergence
edited Dec 22 '18 at 14:23
Paul Frost
10.3k3933
asked Aug 12 '13 at 4:28
MaryamMaryam
311
$endgroup$
add a comment |
$begingroup$
The bellow Theorem is in " Spaces in which compact subsets are close and the lattice of $T_{1}$ topologies on sets" by " T.Alas , G.Wilson"
The one point compactification of a countable $KC$ space$X$ is $KC$ if and only if $X$ is compact.
proof:
sufficiency: an open subspace of a sequential spaces is sequential.
necessity: supposr that $C$ is a compact subspace of the one point compatification $ Y = X cup { infty }$of $X$. If $ infty notin C $ then $C$ is a compact subspace of $X$ , hence closed and so $ Y - C $ is open in $Y$.
if on the other hand $infty in C $, then if $C$ is notclosed in $Y$, $ C cap X$ is not closed in$X$ and hence there is a sequence ${ x_{n} } $in $C cap X $conerging to some $ p notin C $. Since $X$ is $KC$, the compact set$ S = {p} cup { x_{n} : n in omega } $is closed in $X$ and so $ Y- S $ is a neighborhood of $infty $ and so $infty $ is not accumulation point of${ x_{n} } $, implying that $C$ is not compact, a contradiction.
a): why in the sufficiency part, $X$ is open and he said an open subspace of a sequential spaces is sequential?
b): why in the last line of necessity part, we say that $C$ is not compact?
(c): In The one point compatification, $infty notin X $ but in the proof said that "if $infty in C$? Why?
real-analysis general-topology convergence
edited Dec 22 '18 at 14:23
Paul Frost
10.3k3933
asked Aug 12 '13 at 4:28
MaryamMaryam
311
$endgroup$
add a comment |
$begingroup$
The bellow Theorem is in " Spaces in which compact subsets are close and the lattice of $T_{1}$ topologies on sets" by " T.Alas , G.Wilson"
The one point compactification of a countable $KC$ space$X$ is $KC$ if and only if $X$ is compact.
proof:
sufficiency: an open subspace of a sequential spaces is sequential.
necessity: supposr that $C$ is a compact subspace of the one point compatification $ Y = X cup { infty }$of $X$. If $ infty notin C $ then $C$ is a compact subspace of $X$ , hence closed and so $ Y - C $ is open in $Y$.
if on the other hand $infty in C $, then if $C$ is notclosed in $Y$, $ C cap X$ is not closed in$X$ and hence there is a sequence ${ x_{n} } $in $C cap X $conerging to some $ p notin C $. Since $X$ is $KC$, the compact set$ S = {p} cup { x_{n} : n in omega } $is closed in $X$ and so $ Y- S $ is a neighborhood of $infty $ and so $infty $ is not accumulation point of${ x_{n} } $, implying that $C$ is not compact, a contradiction.
a): why in the sufficiency part, $X$ is open and he said an open subspace of a sequential spaces is sequential?
b): why in the last line of necessity part, we say that $C$ is not compact?
(c): In The one point compatification, $infty notin X $ but in the proof said that "if $infty in C$? Why?
real-analysis general-topology convergence
edited Dec 22 '18 at 14:23
Paul Frost
10.3k3933
asked Aug 12 '13 at 4:28
MaryamMaryam
311
$endgroup$
The bellow Theorem is in " Spaces in which compact subsets are close and the lattice of $T_{1}$ topologies on sets" by " T.Alas , G.Wilson"
The one point compactification of a countable $KC$ space$X$ is $KC$ if and only if $X$ is compact.
proof:
sufficiency: an open subspace of a sequential spaces is sequential.
necessity: supposr that $C$ is a compact subspace of the one point compatification $ Y = X cup { infty }$of $X$. If $ infty notin C $ then $C$ is a compact subspace of $X$ , hence closed and so $ Y - C $ is open in $Y$.
if on the other hand $infty in C $, then if $C$ is notclosed in $Y$, $ C cap X$ is not closed in$X$ and hence there is a sequence ${ x_{n} } $in $C cap X $conerging to some $ p notin C $. Since $X$ is $KC$, the compact set$ S = {p} cup { x_{n} : n in omega } $is closed in $X$ and so $ Y- S $ is a neighborhood of $infty $ and so $infty $ is not accumulation point of${ x_{n} } $, implying that $C$ is not compact, a contradiction.
a): why in the sufficiency part, $X$ is open and he said an open subspace of a sequential spaces is sequential?
b): why in the last line of necessity part, we say that $C$ is not compact?
(c): In The one point compatification, $infty notin X $ but in the proof said that "if $infty in C$? Why?
real-analysis general-topology convergence
real-analysis general-topology convergence
edited Dec 22 '18 at 14:23
Paul Frost
10.3k3933
asked Aug 12 '13 at 4:28
MaryamMaryam
311
edited Dec 22 '18 at 14:23
Paul Frost
10.3k3933
asked Aug 12 '13 at 4:28
MaryamMaryam
311
edited Dec 22 '18 at 14:23
Paul Frost
10.3k3933
edited Dec 22 '18 at 14:23
Paul Frost
10.3k3933
edited Dec 22 '18 at 14:23
Paul Frost
10.3k3933
10.3k3933
asked Aug 12 '13 at 4:28
MaryamMaryam
311
asked Aug 12 '13 at 4:28
MaryamMaryam
311
asked Aug 12 '13 at 4:28
MaryamMaryam
311
311
add a comment |
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You stated the theorem incorrectly. It should read:
The one-point compactification of a countable $KC$ space $X$ is $KC$ if and only if $X$ is sequential.
Let $Y$ be the one-point compactification of $X$, and suppose that $Y$ is $KC$; then $Y$ is sequential by Corollary $3$ of the paper. $Y$ is $T_1$, so $X=Ysetminus{infty}$ is open in $Y$ and is therefore sequential.
Let $A={x_n:ninomega}$. Without loss of generality the map $omegato A:nmapsto x_n$ is injective, so for each $ninomega$ the set $S_n={p}cup{x_k:kge n}$ is compact and therefore closed in $X$. Let $ninomega$. $X$ is $T_1$, so $x_n$ has an open nbhd $U_n$ in $X$ such that $x_knotin U_n$ if $k<n$. Let $V_n=U_nsetminus S_{n+1}$; then $V_n$ is an open nbhd of $x_n$, and $V_ncap A={x_n}$. Thus, $A$ is a discrete set in $X$, and since $infty$ is not an accumulation point of $A$ in $Y$, $A$ is actually an infinite discrete set in $Y$. $A$ has no accumulation point in $C$, so $A$ is an infinite, closed, discrete subset of $C$, and hence $C$ cannot be compact.
$C$ is a subset of $Y$, not $X$, so it's entirely possible that $inftyin C$.
answered Aug 12 '13 at 16:39
Brian M. ScottBrian M. Scott
457k38509909
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$begingroup$
You stated the theorem incorrectly. It should read:
The one-point compactification of a countable $KC$ space $X$ is $KC$ if and only if $X$ is sequential.
Let $Y$ be the one-point compactification of $X$, and suppose that $Y$ is $KC$; then $Y$ is sequential by Corollary $3$ of the paper. $Y$ is $T_1$, so $X=Ysetminus{infty}$ is open in $Y$ and is therefore sequential.
Let $A={x_n:ninomega}$. Without loss of generality the map $omegato A:nmapsto x_n$ is injective, so for each $ninomega$ the set $S_n={p}cup{x_k:kge n}$ is compact and therefore closed in $X$. Let $ninomega$. $X$ is $T_1$, so $x_n$ has an open nbhd $U_n$ in $X$ such that $x_knotin U_n$ if $k<n$. Let $V_n=U_nsetminus S_{n+1}$; then $V_n$ is an open nbhd of $x_n$, and $V_ncap A={x_n}$. Thus, $A$ is a discrete set in $X$, and since $infty$ is not an accumulation point of $A$ in $Y$, $A$ is actually an infinite discrete set in $Y$. $A$ has no accumulation point in $C$, so $A$ is an infinite, closed, discrete subset of $C$, and hence $C$ cannot be compact.
$C$ is a subset of $Y$, not $X$, so it's entirely possible that $inftyin C$.
answered Aug 12 '13 at 16:39
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
add a comment |
$begingroup$
You stated the theorem incorrectly. It should read:
The one-point compactification of a countable $KC$ space $X$ is $KC$ if and only if $X$ is sequential.
Let $Y$ be the one-point compactification of $X$, and suppose that $Y$ is $KC$; then $Y$ is sequential by Corollary $3$ of the paper. $Y$ is $T_1$, so $X=Ysetminus{infty}$ is open in $Y$ and is therefore sequential.
Let $A={x_n:ninomega}$. Without loss of generality the map $omegato A:nmapsto x_n$ is injective, so for each $ninomega$ the set $S_n={p}cup{x_k:kge n}$ is compact and therefore closed in $X$. Let $ninomega$. $X$ is $T_1$, so $x_n$ has an open nbhd $U_n$ in $X$ such that $x_knotin U_n$ if $k<n$. Let $V_n=U_nsetminus S_{n+1}$; then $V_n$ is an open nbhd of $x_n$, and $V_ncap A={x_n}$. Thus, $A$ is a discrete set in $X$, and since $infty$ is not an accumulation point of $A$ in $Y$, $A$ is actually an infinite discrete set in $Y$. $A$ has no accumulation point in $C$, so $A$ is an infinite, closed, discrete subset of $C$, and hence $C$ cannot be compact.
$C$ is a subset of $Y$, not $X$, so it's entirely possible that $inftyin C$.
answered Aug 12 '13 at 16:39
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
add a comment |
$begingroup$
You stated the theorem incorrectly. It should read:
The one-point compactification of a countable $KC$ space $X$ is $KC$ if and only if $X$ is sequential.
Let $Y$ be the one-point compactification of $X$, and suppose that $Y$ is $KC$; then $Y$ is sequential by Corollary $3$ of the paper. $Y$ is $T_1$, so $X=Ysetminus{infty}$ is open in $Y$ and is therefore sequential.
Let $A={x_n:ninomega}$. Without loss of generality the map $omegato A:nmapsto x_n$ is injective, so for each $ninomega$ the set $S_n={p}cup{x_k:kge n}$ is compact and therefore closed in $X$. Let $ninomega$. $X$ is $T_1$, so $x_n$ has an open nbhd $U_n$ in $X$ such that $x_knotin U_n$ if $k<n$. Let $V_n=U_nsetminus S_{n+1}$; then $V_n$ is an open nbhd of $x_n$, and $V_ncap A={x_n}$. Thus, $A$ is a discrete set in $X$, and since $infty$ is not an accumulation point of $A$ in $Y$, $A$ is actually an infinite discrete set in $Y$. $A$ has no accumulation point in $C$, so $A$ is an infinite, closed, discrete subset of $C$, and hence $C$ cannot be compact.
$C$ is a subset of $Y$, not $X$, so it's entirely possible that $inftyin C$.
answered Aug 12 '13 at 16:39
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
You stated the theorem incorrectly. It should read:
The one-point compactification of a countable $KC$ space $X$ is $KC$ if and only if $X$ is sequential.
Let $Y$ be the one-point compactification of $X$, and suppose that $Y$ is $KC$; then $Y$ is sequential by Corollary $3$ of the paper. $Y$ is $T_1$, so $X=Ysetminus{infty}$ is open in $Y$ and is therefore sequential.
Let $A={x_n:ninomega}$. Without loss of generality the map $omegato A:nmapsto x_n$ is injective, so for each $ninomega$ the set $S_n={p}cup{x_k:kge n}$ is compact and therefore closed in $X$. Let $ninomega$. $X$ is $T_1$, so $x_n$ has an open nbhd $U_n$ in $X$ such that $x_knotin U_n$ if $k<n$. Let $V_n=U_nsetminus S_{n+1}$; then $V_n$ is an open nbhd of $x_n$, and $V_ncap A={x_n}$. Thus, $A$ is a discrete set in $X$, and since $infty$ is not an accumulation point of $A$ in $Y$, $A$ is actually an infinite discrete set in $Y$. $A$ has no accumulation point in $C$, so $A$ is an infinite, closed, discrete subset of $C$, and hence $C$ cannot be compact.
$C$ is a subset of $Y$, not $X$, so it's entirely possible that $inftyin C$.
answered Aug 12 '13 at 16:39
Brian M. ScottBrian M. Scott
457k38509909
answered Aug 12 '13 at 16:39
Brian M. ScottBrian M. Scott
457k38509909
answered Aug 12 '13 at 16:39
Brian M. ScottBrian M. Scott
457k38509909
answered Aug 12 '13 at 16:39
Brian M. ScottBrian M. Scott
457k38509909
457k38509909
add a comment |
add a comment |
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