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I would like to ask if the following statement is true and, if it is, how could we prove it:

Let $(X,\mathcal T), (X,\mathcal T')$ two homeomorphic topological spaces on the same non empty set $X$. Then $\mathcal T=\mathcal T'$.

If it is false, then how do we say that a metrizable topological space (i.e. a topological space homeomorphic to a metric space) preserves the topology (i.e. the $\mathcal T$-open and $\mathcal T _d$-open sets coincide, where $(X, \mathcal T), (X,d)$ the two spaces)?

Thanks a lot!

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  • $\begingroup$ What means to be homeomorphic in terms of the existence of continuous maps? What can you say about the inverse image of an open subset by a continuous map? $\endgroup$ Apr 12 at 9:37
  • $\begingroup$ The inverse image of an open set under a homeomorphism is an open set. Sorry, I do not understand the purpose of your question. $\endgroup$
    – SK_
    Apr 12 at 9:39
  • $\begingroup$ Your 2nd question is also unclear. A topological space $(X, \mathcal T)$ is said to be metrizable iff there is a distance $d$ on $X$ such that $ \mathcal T_d= \mathcal T$. $\endgroup$ Apr 12 at 9:39
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    $\begingroup$ Yes indeed, for the second question I guess this is not really a question since both possible definition (the one by @AnneBauval and the one by you) give the same result: $\mathcal{T} = \mathcal{T}_d$ for the definition of Anne, but in the second definition the "implicitly" defined metric is given by $\phi:X \to Y$ homeo and $d_X(x,x') = d_Y(\phi(x),\phi(x'))$ so once again by definition we would have $\mathcal{T} = \mathcal{T}_{d_X}$ $\endgroup$ Apr 12 at 10:18
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    $\begingroup$ In general if $\mathcal T$ is a topology which is a non-counterexample - so is invariant under permutations - then whenever $U$ is open and we have $|U| = |V|$ and $|X \setminus U| = |X \setminus V|$, then $V$ must also be open. So any topology in which openness depends on something more than cardinality must be a counterexample - this is almost all "nice" topologies. I believe it follows that the only non-counterexamples are the co-$\kappa$ spaces - eg the discrete space, the indiscrete space, the cofinite space, the cocountable space, etc. $\endgroup$ Apr 12 at 10:40

5 Answers 5

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Here is the counterexample on $X=\{0,1\}$ suggested in my earlier comment:

$$\mathcal T:=\{\varnothing,\{0\},X\}\ne\mathcal T':=\{\varnothing,\{1\},X\}$$

although the transposition $0\mapsto1,1\mapsto0$ is a homeomorphism between $(X,\mathcal T)$ and $(X,\mathcal T')$.

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To answer your title question: Yes, a homemorphism does preserve open sets. If $\newcommand{\T}{\mathcal{T}}(X,\T)$ and $(X',\T')$ are homeomorphic under $f : X \to X'$, then modulo this bijection $f$, $\T$ corresponds to $\T'$ — that is, $\T = \{ f^{-1}(U) \mid U \in \T' \}$, or equivalently, $\T' = \{ f(U) \mid U \in \T \}$.

However, this doesn’t imply the statement you give in the body, that if $(X,\T)$ and $(X,\T')$ are homeomorphic then $\T = \T'$. Being homeomorphic means there’s a homeorphism $f : X \to X$, and modulo this bijection $f$, $\T$ corresponds to $\T'$ in the sense above. But this doesn’t imply $\T = \T'$, unless $f$ is the identity map on $X$.

Other answers give nice finite counterexamples. Another example, natural and fairly intuitive, is $\newcommand{\R}{\mathbb{R}}\R$ with the upper or lower interval topologies (generated by intervals $(x,\infty)$ for the upper case, or alternatively $(-\infty,x)$ for the lower). These are not the same topology — but the only difference is a reversal of the order, so they’re homeomorphic to each other via the order-reversal map $x \mapsto -x$.

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Let $X = \{1,2,3\}$ and let $\mathcal T = \{ \emptyset, \{1\}, \{2,3\}, X\}$ and $\mathcal T' = \{ \emptyset, \{2\}, \{1,3\}, X\}$. These spaces are homeomorphic via the map $1 \mapsto 2, 2 \mapsto 1, 3 \mapsto 3$ but $\mathcal T \neq \mathcal T'$.

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An easy counterexample with metric spaces: for $a\in\mathbb{R}$, define the following metric $$ d_a(x,y)=\begin{cases} \min\{|x-y|,1\} & x\ne a,y\ne a \\[1ex] 1 & x=a,y\ne a \\[1ex] 1 & x\ne a,y=a \\[1ex] 0 & x=y=a \end{cases} $$ The map $f\colon(\mathbb{R},d_0)\to (\mathbb{R},d_1)$ defined by $f(x)=x+1$ is a homeomorphism.

The set $\{0\}$ is open for the topology induced by $d_0$, but it isn't open for the topology induced by $d_1$.

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I would like to summarize somehow.

Suppose the topological spaces $(X,\mathcal T),(Y,\mathcal T')$, where $X,Y$ nonempty sets. If they are homeomorphic, i.e. there exists a bicontinuous map $\phi:(X,\mathcal T)\to(Y,\mathcal T')$, then the open sets are preserved:

We have $G\in \mathcal T'\implies \phi^{-1}(G):=U\in \mathcal T\implies G=\phi (U)\in \mathcal T'$, for $\phi$ is a homeomorphism, thus it is an open map and its inverse is an open map too. Hence $\mathcal T'=\{\phi (U)|U\in \mathcal T\}$. Similarly $\mathcal T=\{\phi^{-1}(V)|V\in \mathcal T'\}$.

Let $(X,\mathcal T)\simeq (X, \mathcal T')$. These homeomorphic topological spaces on the same underlying set $X$ do not need to share the same topology (i.e. $\mathcal T=\mathcal T'$). As a counterexample we can consider $\mathcal T, \mathcal T'$ as the Sierpiński topologies with their singleton different.

Finally, if $(X, \mathcal T)$ is metrizable, i.e. there is a metric $d:X^2\to \Bbb R$ such that $\mathcal T=\mathcal T_d$, then $(X, \mathcal T),(X,\mathcal T_d)$ are homeomorphic under the identity map $\rm id_X$$:X\to X$. Finally, if $(X, \mathcal T)\simeq (X,\mathcal T_d)$ for some metric $d:X^2\to\Bbb R$, then the topological space is metrizable:

Consider the metric $\rho (x,y):=d(\phi(x), \phi(y))$ for $x,y\in X$. Use the equality $\phi (B_{\rho}(x,\epsilon))=B_d(\phi (x), \epsilon)$ to prove that $U\in \mathcal T_{\rho}\iff U\in \mathcal T$.

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  • $\begingroup$ This summary looks almost right. But I think that if $(X, \mathcal T)$ is homeomorphic to $(X, \mathcal T_d)$ for some metric topology $\mathcal T_d$, then it is necessarily the case that $\mathcal T$ is equal to the metric topology $\mathcal T_{d'}$ for some metric $d'$. The point is that you "transport" $d$ along the earlier homeomorphism. That's the content of this comment. The Sierpinski space isn't homeomorphic to any metric topology because it's not Hausdorff. $\endgroup$ Apr 15 at 12:00
  • $\begingroup$ You are right about the second counterexample (Sierpiński space isn' t metrizable); it is my fault and I am going to remove it. $\endgroup$
    – SK_
    Apr 16 at 16:15

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