Can a noncompact metric space have a maximal metrizable Hausdorff compactification?
We know the Stone-Čech compactification $(h, beta X)$ of a Tychonoff space $X$ is its largest (in particular, a maximal) Hausdorff compactification, in the sense that if $(k,gamma X$) is any other Hausdorff compactification, then there is a continuous map $f:beta X
rightarrow gamma X$ such that $fh=k$.
What about if I restrict to considering just all metrizable Hausdorff compactifications? To fix ideas, say $X$ is a noncompact metric space. (In particular $beta X$ will not be metrizable.)
Is there a maximal metrizable Hausdorff compactification of $X$?
I am guessing no: if we're given any metrizable Hausdorff compactification of noncompact metric space $X$, we can make another metrizable Hausdorff compactification that is larger (in the sense of the first paragraph above).
general-topology metric-spaces compactness compactification
edited Nov 28 '18 at 21:06
Eric Wofsey
180k12207335
asked Nov 28 '18 at 20:02
SSF
362110
add a comment |
We know the Stone-Čech compactification $(h, beta X)$ of a Tychonoff space $X$ is its largest (in particular, a maximal) Hausdorff compactification, in the sense that if $(k,gamma X$) is any other Hausdorff compactification, then there is a continuous map $f:beta X
rightarrow gamma X$ such that $fh=k$.
What about if I restrict to considering just all metrizable Hausdorff compactifications? To fix ideas, say $X$ is a noncompact metric space. (In particular $beta X$ will not be metrizable.)
Is there a maximal metrizable Hausdorff compactification of $X$?
I am guessing no: if we're given any metrizable Hausdorff compactification of noncompact metric space $X$, we can make another metrizable Hausdorff compactification that is larger (in the sense of the first paragraph above).
general-topology metric-spaces compactness compactification
edited Nov 28 '18 at 21:06
Eric Wofsey
180k12207335
asked Nov 28 '18 at 20:02
SSF
362110
add a comment |
We know the Stone-Čech compactification $(h, beta X)$ of a Tychonoff space $X$ is its largest (in particular, a maximal) Hausdorff compactification, in the sense that if $(k,gamma X$) is any other Hausdorff compactification, then there is a continuous map $f:beta X
rightarrow gamma X$ such that $fh=k$.
What about if I restrict to considering just all metrizable Hausdorff compactifications? To fix ideas, say $X$ is a noncompact metric space. (In particular $beta X$ will not be metrizable.)
Is there a maximal metrizable Hausdorff compactification of $X$?
I am guessing no: if we're given any metrizable Hausdorff compactification of noncompact metric space $X$, we can make another metrizable Hausdorff compactification that is larger (in the sense of the first paragraph above).
general-topology metric-spaces compactness compactification
edited Nov 28 '18 at 21:06
Eric Wofsey
180k12207335
asked Nov 28 '18 at 20:02
SSF
362110
We know the Stone-Čech compactification $(h, beta X)$ of a Tychonoff space $X$ is its largest (in particular, a maximal) Hausdorff compactification, in the sense that if $(k,gamma X$) is any other Hausdorff compactification, then there is a continuous map $f:beta X
rightarrow gamma X$ such that $fh=k$.
What about if I restrict to considering just all metrizable Hausdorff compactifications? To fix ideas, say $X$ is a noncompact metric space. (In particular $beta X$ will not be metrizable.)
Is there a maximal metrizable Hausdorff compactification of $X$?
I am guessing no: if we're given any metrizable Hausdorff compactification of noncompact metric space $X$, we can make another metrizable Hausdorff compactification that is larger (in the sense of the first paragraph above).
general-topology metric-spaces compactness compactification
general-topology metric-spaces compactness compactification
edited Nov 28 '18 at 21:06
Eric Wofsey
180k12207335
asked Nov 28 '18 at 20:02
SSF
362110
edited Nov 28 '18 at 21:06
Eric Wofsey
180k12207335
asked Nov 28 '18 at 20:02
SSF
362110
edited Nov 28 '18 at 21:06
Eric Wofsey
180k12207335
edited Nov 28 '18 at 21:06
Eric Wofsey
180k12207335
edited Nov 28 '18 at 21:06
Eric Wofsey
180k12207335
180k12207335
asked Nov 28 '18 at 20:02
SSF
362110
asked Nov 28 '18 at 20:02
SSF
362110
asked Nov 28 '18 at 20:02
SSF
362110
362110
add a comment |
add a comment |
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No. Let $X$ be a noncompact metric space and let $Y$ be a metrizable compactification of $X$. Pick a point $yin Y$ and a sequence $(x_n)$ of distinct points in $X$ converging to $y$. Let $A$ consist of the points $x_n$ for all even $n$ and let $B$ consist of the points $x_n$ for all odd $n$. Then $A$ and $B$ are closed and disjoint in $X$, so there exists a continuous $f:Xto[0,1]$ such that $f$ maps $A$ to $0$ and $B$ to $1$. This map $f$ does not extend continuously to $Y$.
We can now embed $X$ in $Ytimes [0,1]$ by using the given embedding $Xto Y$ on the first coordinate and $f$ on the second coordinate. The closure of the image of $X$ in $Ytimes[0,1]$ is then a metrizable compactification of $X$, which is strictly larger than $Y$ since $f$ extends continuously to it.
For another perspective on this, note that the algebra $C_b(X)$ of bounded continuous functions on $X$ can canonically be identified with $C(beta X)$, and an arbitrary compactification of $X$ just corresponds to a closed subalgebra of $C_b(X)$ which separates points from closed sets on $X$. Moreover, a compactification is metrizable iff the corresponding subalgebra is separable. "Larger" compactifications in your sense just corresponds to the inclusion order on subalgebras.
So, given any metrizable compactification corresponding to a separable subalgebra $C(Y)subset C_b(X)$, you can get a larger separable subalgebra by just taking some $fin C_b(X)setminus C(Y)$ and taking the closed subalgebra generated by $f$ and $C(Y)$.
answered Nov 28 '18 at 21:01
Eric Wofsey
180k12207335
May I ask why the closure of the image of $X$ in $Y×[0,1]$ is a compactification of $X$? I've gone through the other details, but in particular why is the inverse function of the embedding continuous?
– SSF
Dec 8 '18 at 2:20
1
The inverse function is the composition of the projection $Ytimes [0,1]to Y$ and the inverse of the embedding $Xto Y$.
– Eric Wofsey
Dec 8 '18 at 2:24
add a comment |
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No. Let $X$ be a noncompact metric space and let $Y$ be a metrizable compactification of $X$. Pick a point $yin Y$ and a sequence $(x_n)$ of distinct points in $X$ converging to $y$. Let $A$ consist of the points $x_n$ for all even $n$ and let $B$ consist of the points $x_n$ for all odd $n$. Then $A$ and $B$ are closed and disjoint in $X$, so there exists a continuous $f:Xto[0,1]$ such that $f$ maps $A$ to $0$ and $B$ to $1$. This map $f$ does not extend continuously to $Y$.
We can now embed $X$ in $Ytimes [0,1]$ by using the given embedding $Xto Y$ on the first coordinate and $f$ on the second coordinate. The closure of the image of $X$ in $Ytimes[0,1]$ is then a metrizable compactification of $X$, which is strictly larger than $Y$ since $f$ extends continuously to it.
For another perspective on this, note that the algebra $C_b(X)$ of bounded continuous functions on $X$ can canonically be identified with $C(beta X)$, and an arbitrary compactification of $X$ just corresponds to a closed subalgebra of $C_b(X)$ which separates points from closed sets on $X$. Moreover, a compactification is metrizable iff the corresponding subalgebra is separable. "Larger" compactifications in your sense just corresponds to the inclusion order on subalgebras.
So, given any metrizable compactification corresponding to a separable subalgebra $C(Y)subset C_b(X)$, you can get a larger separable subalgebra by just taking some $fin C_b(X)setminus C(Y)$ and taking the closed subalgebra generated by $f$ and $C(Y)$.
answered Nov 28 '18 at 21:01
Eric Wofsey
180k12207335
May I ask why the closure of the image of $X$ in $Y×[0,1]$ is a compactification of $X$? I've gone through the other details, but in particular why is the inverse function of the embedding continuous?
– SSF
Dec 8 '18 at 2:20
1
The inverse function is the composition of the projection $Ytimes [0,1]to Y$ and the inverse of the embedding $Xto Y$.
– Eric Wofsey
Dec 8 '18 at 2:24
add a comment |
No. Let $X$ be a noncompact metric space and let $Y$ be a metrizable compactification of $X$. Pick a point $yin Y$ and a sequence $(x_n)$ of distinct points in $X$ converging to $y$. Let $A$ consist of the points $x_n$ for all even $n$ and let $B$ consist of the points $x_n$ for all odd $n$. Then $A$ and $B$ are closed and disjoint in $X$, so there exists a continuous $f:Xto[0,1]$ such that $f$ maps $A$ to $0$ and $B$ to $1$. This map $f$ does not extend continuously to $Y$.
We can now embed $X$ in $Ytimes [0,1]$ by using the given embedding $Xto Y$ on the first coordinate and $f$ on the second coordinate. The closure of the image of $X$ in $Ytimes[0,1]$ is then a metrizable compactification of $X$, which is strictly larger than $Y$ since $f$ extends continuously to it.
For another perspective on this, note that the algebra $C_b(X)$ of bounded continuous functions on $X$ can canonically be identified with $C(beta X)$, and an arbitrary compactification of $X$ just corresponds to a closed subalgebra of $C_b(X)$ which separates points from closed sets on $X$. Moreover, a compactification is metrizable iff the corresponding subalgebra is separable. "Larger" compactifications in your sense just corresponds to the inclusion order on subalgebras.
So, given any metrizable compactification corresponding to a separable subalgebra $C(Y)subset C_b(X)$, you can get a larger separable subalgebra by just taking some $fin C_b(X)setminus C(Y)$ and taking the closed subalgebra generated by $f$ and $C(Y)$.
answered Nov 28 '18 at 21:01
Eric Wofsey
180k12207335
May I ask why the closure of the image of $X$ in $Y×[0,1]$ is a compactification of $X$? I've gone through the other details, but in particular why is the inverse function of the embedding continuous?
– SSF
Dec 8 '18 at 2:20
1
The inverse function is the composition of the projection $Ytimes [0,1]to Y$ and the inverse of the embedding $Xto Y$.
– Eric Wofsey
Dec 8 '18 at 2:24
add a comment |
No. Let $X$ be a noncompact metric space and let $Y$ be a metrizable compactification of $X$. Pick a point $yin Y$ and a sequence $(x_n)$ of distinct points in $X$ converging to $y$. Let $A$ consist of the points $x_n$ for all even $n$ and let $B$ consist of the points $x_n$ for all odd $n$. Then $A$ and $B$ are closed and disjoint in $X$, so there exists a continuous $f:Xto[0,1]$ such that $f$ maps $A$ to $0$ and $B$ to $1$. This map $f$ does not extend continuously to $Y$.
We can now embed $X$ in $Ytimes [0,1]$ by using the given embedding $Xto Y$ on the first coordinate and $f$ on the second coordinate. The closure of the image of $X$ in $Ytimes[0,1]$ is then a metrizable compactification of $X$, which is strictly larger than $Y$ since $f$ extends continuously to it.
For another perspective on this, note that the algebra $C_b(X)$ of bounded continuous functions on $X$ can canonically be identified with $C(beta X)$, and an arbitrary compactification of $X$ just corresponds to a closed subalgebra of $C_b(X)$ which separates points from closed sets on $X$. Moreover, a compactification is metrizable iff the corresponding subalgebra is separable. "Larger" compactifications in your sense just corresponds to the inclusion order on subalgebras.
So, given any metrizable compactification corresponding to a separable subalgebra $C(Y)subset C_b(X)$, you can get a larger separable subalgebra by just taking some $fin C_b(X)setminus C(Y)$ and taking the closed subalgebra generated by $f$ and $C(Y)$.
answered Nov 28 '18 at 21:01
Eric Wofsey
180k12207335
No. Let $X$ be a noncompact metric space and let $Y$ be a metrizable compactification of $X$. Pick a point $yin Y$ and a sequence $(x_n)$ of distinct points in $X$ converging to $y$. Let $A$ consist of the points $x_n$ for all even $n$ and let $B$ consist of the points $x_n$ for all odd $n$. Then $A$ and $B$ are closed and disjoint in $X$, so there exists a continuous $f:Xto[0,1]$ such that $f$ maps $A$ to $0$ and $B$ to $1$. This map $f$ does not extend continuously to $Y$.
We can now embed $X$ in $Ytimes [0,1]$ by using the given embedding $Xto Y$ on the first coordinate and $f$ on the second coordinate. The closure of the image of $X$ in $Ytimes[0,1]$ is then a metrizable compactification of $X$, which is strictly larger than $Y$ since $f$ extends continuously to it.
For another perspective on this, note that the algebra $C_b(X)$ of bounded continuous functions on $X$ can canonically be identified with $C(beta X)$, and an arbitrary compactification of $X$ just corresponds to a closed subalgebra of $C_b(X)$ which separates points from closed sets on $X$. Moreover, a compactification is metrizable iff the corresponding subalgebra is separable. "Larger" compactifications in your sense just corresponds to the inclusion order on subalgebras.
So, given any metrizable compactification corresponding to a separable subalgebra $C(Y)subset C_b(X)$, you can get a larger separable subalgebra by just taking some $fin C_b(X)setminus C(Y)$ and taking the closed subalgebra generated by $f$ and $C(Y)$.
answered Nov 28 '18 at 21:01
Eric Wofsey
180k12207335
edited Nov 28 '18 at 21:06
answered Nov 28 '18 at 21:01
Eric Wofsey
180k12207335
answered Nov 28 '18 at 21:01
Eric Wofsey
180k12207335
answered Nov 28 '18 at 21:01
Eric Wofsey
180k12207335
180k12207335
May I ask why the closure of the image of $X$ in $Y×[0,1]$ is a compactification of $X$? I've gone through the other details, but in particular why is the inverse function of the embedding continuous?
– SSF
Dec 8 '18 at 2:20
1
The inverse function is the composition of the projection $Ytimes [0,1]to Y$ and the inverse of the embedding $Xto Y$.
– Eric Wofsey
Dec 8 '18 at 2:24
add a comment |
May I ask why the closure of the image of $X$ in $Y×[0,1]$ is a compactification of $X$? I've gone through the other details, but in particular why is the inverse function of the embedding continuous?
– SSF
Dec 8 '18 at 2:20
1
The inverse function is the composition of the projection $Ytimes [0,1]to Y$ and the inverse of the embedding $Xto Y$.
– Eric Wofsey
Dec 8 '18 at 2:24
May I ask why the closure of the image of $X$ in $Y×[0,1]$ is a compactification of $X$? I've gone through the other details, but in particular why is the inverse function of the embedding continuous?
– SSF
Dec 8 '18 at 2:20
May I ask why the closure of the image of $X$ in $Y×[0,1]$ is a compactification of $X$? I've gone through the other details, but in particular why is the inverse function of the embedding continuous?
– SSF
Dec 8 '18 at 2:20
1
1
The inverse function is the composition of the projection $Ytimes [0,1]to Y$ and the inverse of the embedding $Xto Y$.
– Eric Wofsey
Dec 8 '18 at 2:24
The inverse function is the composition of the projection $Ytimes [0,1]to Y$ and the inverse of the embedding $Xto Y$.
– Eric Wofsey
Dec 8 '18 at 2:24
add a comment |
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