# Does a homeomorphism preserve the open sets?

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!

• 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? Apr 12 at 9:37
• The inverse image of an open set under a homeomorphism is an open set. Sorry, I do not understand the purpose of your question.
– SK_
Apr 12 at 9:39
• 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 $\mathcal T_d= \mathcal T$. Apr 12 at 9:39
• 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 $\mathcal{T} = \mathcal{T}_{d_X}$ Apr 12 at 10:18
• 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. Apr 12 at 10:40

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 $$(X,\mathcal T')$$.

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 $$x \mapsto -x$$.

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 $$\mathcal T \neq \mathcal T'$$.

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 $$d_1$$.

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 $$U\in \mathcal T_{\rho}\iff U\in \mathcal T$$.

• 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. Apr 15 at 12:00
• You are right about the second counterexample (Sierpiński space isn' t metrizable); it is my fault and I am going to remove it.
– SK_
Apr 16 at 16:15