Does a homeomorphism preserve the open sets?

Question

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!

Answer 1

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 $(X,\mathcal T')$.

Answer 2

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 $x \mapsto -x$.

Answer 3

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

Answer 4

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 $d_1$.

Answer 5

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