9

Let $X$ and $Y$ be topological spaces and let $A \subseteq X$ be a subspace of $X$. Suppose $A$ is homeomorphic to some subspace $B \subseteq Y$ of $Y$. Let $f$ explicitly denote this homeomorphism.

If $f : A \to B$ is a homeomorphism, does $f$ extend to a homeomorphism between $\text{Cl}_X(A)$ and $\text{Cl}_Y(B)$, i.e does there exist a $g : Cl_X(A) \to Cl_Y(B)$, such that $g|_{A} = f$.

More generally if $A$ and $B$ are homeomorphic, does that imply that $Cl_X(A)$ and $Cl_Y(B)$ are homeomorphic?

If $X$ and $Y$ are homeormorphic, then this is true, since it is a well known-theorem that $f[Cl_X(A)] = Cl_Y(f[A] = B)$, if $f : X \to Y$ is a homeomorphism.

However I can't seem to come up with a counterexample for my question, since I assume the implication is false.

Edit : I know a counterexample that comes from CW Complexes, where if $(X, \xi)$ is a CW-Complex, then $\xi$ is a collection of open cells $e$, which are topological spaces homeomorphic to $\mathbb{B}^n$ the open unit ball in $\mathbb{R}^n$. Each $e \subseteq X$ is a subspace of a haursdoff space $X$.

Also a closed cell $\bar{e}$ is a topological space homeomorphic to the closed unit ball $\mathbb{\bar{B}}^n \subseteq \mathbb{R}^n$

Now in this example $Y = \mathbb{R}^n$. It is also known that for any $e \in \xi$, $Cl_X(e) \neq \bar{e} \cong {\mathbb{\bar{B}}^n} = Cl_Y(\mathbb{{B}}^n \cong e) $

Hence $e$ and $\mathbb{B}^n$ are homeomorphic, however $Cl_X(e)$ is not homeomorphic to $Cl_Y(\mathbb{B}^n) = \mathbb{\bar{B}}^n$, since $Cl_X(e) \neq \bar{e}$

But using the above example feels like bringing a gun to a knife fight, are there any simpler counterexamples?

Perturbative
  • 12,972

3 Answers3

13

$A=S^1\setminus \{p\} \subseteq X= S^1$ (for any $p \in S^1$) is homeomorphic to $B=(0,1) \subseteq Y = [0,1]$. But their respective closures $X$ and $Y$ are not.

More trivially: $A = (0,1) \subseteq X=\mathbb{R}$ and $B = Y = \mathbb{R}$, where $\overline{B} = B$ but $\overline{A}$ becomes compact.

Henno Brandsma
  • 242,131
4

Let $A$ be a non closed subset of the compact space Hausdorff space $Y$, with the induced topology; now take $X=A$ and $f\colon X\to Y$ the inclusion map. The closure of $A$ in $X$ is $A$, not compact; the closure of $f(A)$ in $Y$ is compact.

Explicit example: $A=X=(0,1)$, $Y=[0,1]$.

egreg
  • 238,574
3

Suppose $X$ has a sub-space $A$ that is homeomorphic to $X$ where $A$ is not a closed subset of $X$ and $Cl_X(A)\ne A$ and where $A$ is not homeomorphic to $Cl_X(A)$.Let $X=Y=B$ and let $f:A\to B$ be any homeomorphism. Since $f$ is a bijection from $A$ onto $Y$ and $Cl_X(A)\ne A,$ therefore $f$ cannot be extended to a bijection from $Cl_X(A)$ onto $Cl_Y(B)=Y.$ And $Cl_Y(B)=Cl_X(X)=X$ is homeomophic to $A$, which is not homeomorphic to $Cl_X(A).$

A simple example is $X=Y=B=\mathbb R$ and $A=(-\pi /2,+\pi /2)$ and $f(x)=\tan x.$

DanielWainfleet
  • 57,985
  • The most elementary way that I know of, to show that $(a,b)$ is not homeomorphic to $[a,b]$ is ( by contradiction) assume $g:(a,b)\to [a,b]$ is a continuous bijection. Let $g(x)=b.$ Let $A=\inf {g(y):a<y<x}.$ Let $B=\inf {g(z):x<z<b}.$ Let $C=\max (A,B).$ Then $C<b,$ so by the Intermediate Value property, there exists $y\in (a,x)$ with $g(y)=(b+C)/2$ and there exists $z\in (x,b)$ with $g(z)=(b+C)/2,$ contradicting the 1-to-1 property of $g.$ – DanielWainfleet Sep 17 '17 at 14:56