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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.

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7 Answers 7

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

To answer your question: yes.

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 $(X,\tau')$.

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  • $\begingroup$ Thanks for your correction. $\endgroup$
    – John
    Commented Feb 4, 2014 at 12:43
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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 $\langle X , \mathcal{O}_f \rangle$.

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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.

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

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  • $\begingroup$ 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. $\endgroup$ Commented Feb 4, 2014 at 22:29
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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 $(\mathbb R,\tau_2)$.

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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.

You asked the question

"is it possible that a set with two different topologies can still be self-homeomorphic?"

and the real aswer here is that if you have a set with two different topologies, you have TWO topological spaces and you cannot even begin to speak of them being self-homeomorphic.

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  • $\begingroup$ I would really like to know what in this answer is false to earn me a downvote. $\endgroup$
    – 5xum
    Commented Feb 4, 2014 at 12:53
  • $\begingroup$ 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 $X$. $\endgroup$ Commented Feb 4, 2014 at 13:36
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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 $E_1$.

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