# Does homeomorphic to itself imply the same topology?

I know homeomorphim is an equivalence relation, which means a topological space will be homeomorphic to itself. However, does the converse hold? In other words, is it possible that a set with two different topologies can still be self-homeomorphic?

Thank you.

Firstly a terminological point: the equivalence relation is called homeomorphic; a "homeomorphism" is a particular kind of function.

In particular, let $X$ denote a set with two or more elements. Then there exist distinct topologies $\tau,\tau' \subseteq \mathcal{P}(X)$ such that $(X,\tau)$ and $(X,\tau')$ are homeomorphic (but distinct).

Proof. Since $X$ has two or more elements, we may let $x$ and $y$ denote distinct points of $X$. Now define $\tau = \{\emptyset,\{x\},X\}$ and $\tau' = \{\emptyset,\{y\},X\}$. Observe that $\tau$ and $\tau'$ are distinct, and that they're both topologies.

Now consider the function $f : X \rightarrow X$ that permutes the points $x$ and $y$ while leaving all the other points constant. This is a homeomorphism $f : (X,\tau) \rightarrow (X,\tau'),$ thus $(X,\tau)$ is homeomorphic to $(X,\tau')$.

– John
Commented Feb 4, 2014 at 12:43

Well... yes. In a bit more generality that has been done so far, suppose $\mathcal{O}$ is a topology on a set $X$, and $f : X \to X$ is a bijection, then $$\mathcal{O}_f := \{ f [ U ] = \{ f(x) : x \in U \} : U \in \mathcal{O} \}$$ is also a topology on $X$. Usually this topology will differ from the original topology (but not always: what if $\mathcal{O}$ is the discrete topology? or if $f$ is the identity function?), but the mapping $f$ will be a homeomorphism from $\langle X , \mathcal{O} \rangle$ onto $\langle X , \mathcal{O}_f \rangle$.

Do you want the homeomorphism to be the identity? In that case, it will follow that the topologies must necessarily agree.

If not, consider $X = \{1,2\}$ with the two (different) topologies $\mathcal{T}_1 = \{\emptyset,\{1\},X\}$ and $\mathcal{T}_2 = \{\emptyset,\{2\},X\}$. The identity is not a homeomorphism in this case but the map interchanging $1$ and $2$ is.

• Thank you for the example.
– John
Commented Feb 4, 2014 at 12:47

Another example: $\mathbb{R}$ in the right-arrow topology (generated by all sets of the form $[a,b)$) is homeomorphic to the $\mathbb{R}$ in the left arrow topology (generated by all sets of the form $(a,b]$), using the homeomorphism $x \rightarrow -x$, but the only topology on the reals having both types of intervals be open is the discrete topology (as $[a,a+1) \cap (a-1, a] = \{a\}$, so singletons in the "union topology" are open). So these topologies on the same set are in a sense "orthogonal" but homeomorphic nonetheless.

• Here there are even quite a lot homeomorphisms. For any pair $y,z$ there is a homeomorphism which sends $y$ to $z$, namely the map $f:x\mapsto-x+y+z$. This is of course because the Sorgenfrey line is a homogeneous space. Commented Feb 4, 2014 at 22:29

To add to the other answers, it is actually possible for a finer topology to be homeomorphic to a coarser one. I.e. you can add open sets to a topology and yet preserve homeomorphism-type.

Example

An example discussed here is $$\mathbb{Q}$$ with the Euclidean topology $$\tau_E$$ and the "Sorgenfrey'' topology $$\tau_S$$ generated by the base $$\{(a,b]\mid a,b\in\mathbb{Q}\}$$ are actually homeomorphic. Clearly, $$(a,b)$$ is always in $$\tau_S$$ but $$(a,b]$$ is never in $$\tau_E$$. So $$\tau_S$$ is strictly finer than $$\tau_E$$

The spaces $$(\mathbb{Q},\tau_E)$$ and $$(\mathbb{Q},\tau_S)$$ are homeomorphic because they are both countable metrizable spaces without isolated points and all such spaces are homeomorphic to $$(\mathbb{Q},\tau_E)$$ by theorem in the link.

Edit (simpler example):

Consider $$\mathbb R$$ with the topologies

$$\tau_1=\{A\mid A^c\text{ is a finite set of integers}\}\cup\{\varnothing\},$$ $$\tau_2=\{A\mid A^c\text{ is a finite set of even integers}\}\cup\{\varnothing\}.$$

Clearly $$\tau_2\subsetneq \tau_1$$ and yet $$x\mapsto 2x$$ is a homeomorphism from $$(\mathbb R,\tau_1)$$ to $$(\mathbb R,\tau_2)$$.

The basic idea you are missing here is this: What type of object does a hoeomorphism apply to? The thing is that you cannot speak of two sets being homeomorphic, only two topological spaces can be homeomorphic.

For example, you can have a topological space $X=\mathbb R$ with the standard topology, and the space $Y=\mathbb R$ with the trivial topology $\{\emptyset, \mathbb R\}$. These two spaces are clearly not homeomorphic.

• I didn't downvote. I think by "self-homeomorphic" the OP meant a homeomorphism $(X,\tau_1)\to(X,\tau_2)$, so the "self" only refers to the set $X$. Commented Feb 4, 2014 at 13:36
Let $E$ be a set equipped with topologies $T_1$ and $T_2$. Then the two spaces $E_1$ and $E_2$ are equivalent iff $\mathrm{id}_E$ is a homeomorphism. Let $E=\boldsymbol R$, $d_1$ the discrete metric and $d_2$ the standard metric. Then the map $x\mapsto x$ from $E_1$ to $E_2$ is continuos, but not from $E_2$ to $E_1$.