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I'm teaching my self topology with the aid of a book, but i'm confused about the meaning of homeomorphic. Below, I have 2 topologies, $\mathscr{T}_1$ and $\mathscr{T}_2$ and I'm pretty sure they are not homeomorphic, but I can make a continuous bijection between them.

Definition of homeomorphism:

If $X$ and $Y$ are topological spaces, a homeomorphism from $X$ to $Y$ is a bijective map $\varphi : X \rightarrow Y$ such that both $\varphi$ and $\varphi^{-1}$ are continuous.

Definition of continuous:

If $X$ and $Y$ are topological spaces, a map $f: X \rightarrow Y$ is said to be continuous if for every open subset $U \subseteq Y$, its preimage $f^{-1}(U)$ is open in $X$.

my (counter) example:

Let: $X = \{ 1,2,3 \}$
$\mathscr{T}_1=\{X,\varnothing,\{1\},\{2\},\{1,2\}\}$;
$\mathscr{T}_2=\{X,\varnothing,\{1\},\{1,2\},\{1,3\}\}$;
$ g: \mathscr{T}_1 \rightarrow \mathscr{T}_2$

$g$ is defined by the table below:

$$\begin{array}{|c|c|} V & g(V) \\ \hline \hline X & X \\ \hline \varnothing & \varnothing \\ \hline \{1\} & \{1\} \\ \hline \{1,2\} & \{1,2\} \\ \hline \{2\} & \{1,3\} \\ \hline \end{array}$$

Then $g^{-1}$ would be:

$$\begin{array}{|c|c|} U & g^{-1}(U) \\ \hline \hline X & X \\ \hline \varnothing & \varnothing \\ \hline \{1\} & \{1\} \\ \hline \{1,2\} & \{1,2\} \\ \hline \{1,3\} & \{2\} \\ \hline \end{array}$$

The space is discrete so all sets are open (and closed). My book says that "size" is not a topological property, so it seems okay to map {2} to {1,3}.

So, it seems to me that $g$ and $g^{-1}$ are continuous and bijective, so the topologies are homeomorphic.

I'm pretty sure i'm doing something wrong, and I have another similar example like this. Please help me understand to meaning of homeomorphic.

note: I Do understand why a line segment and a circle are not homeopathic, when there are two topological spaces generated by different metrics (on the same set), by showing there are sets that are open in one space but not the other. But in this example everything is open, so wouldn't all bijections be continuous in a discrete space?

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  • $\begingroup$ There are two concepts need to clarify, I guess. 1.function. 2.set function. In the definition of continuous, $f$ means the set function of the function $f$. They use the same symbol $f$. $\endgroup$ – yoyo Jun 25 '16 at 14:52
  • $\begingroup$ Are you saying my $g$ is not a proper "set function," If so what would be an example of a proper (set) function in this case? $\endgroup$ – Michael Maliszesky Jun 25 '16 at 14:58
  • $\begingroup$ Excuse me, can you please explain what is "proper" you describe here? $\endgroup$ – yoyo Jun 25 '16 at 15:41
  • $\begingroup$ I just made a follow up post, in it I wrote what I now think is as "proper" function. math.stackexchange.com/questions/1839880/… $\endgroup$ – Michael Maliszesky Jun 26 '16 at 4:19
  • $\begingroup$ I think you are right in the new question. $\endgroup$ – yoyo Jun 26 '16 at 4:41
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Like par said, the mapping is supposed to be on the underlying sets themselves. When we have a continuous function $g: X \rightarrow Y$ where $X$ and $Y$ are topological spaces, the function sends points in the set $X$ to points in in the set $Y$. Notice that if $\{2\}$ is being mapped to $\{1,3\}$, then 2 is mapped to 1 and 3 which can't even happen with a function. Your function is still a bijection between the two topologies but not the sets $X$ and $Y$.

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  • $\begingroup$ Thanks, If the spaces are $(X,\mathscr{T}_1)$ and $(X,\mathscr{T}_2)$. Then couldn't the map simply be the identity map, since they both use the same set? $\endgroup$ – Michael Maliszesky Jun 25 '16 at 15:12
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    $\begingroup$ If that was the case, then for example the inverse of {1,3} would still be {1,3}, but {1,3} is not open in the first topology. Basically two spaces are homeomorphic if they are really the same space only represented differently. It's like how a group isomorphism is a map between two groups that are actually the same group only represented with different symbols. The two topological spaces you have defined are not homeomorphic. You have one of them have two one point sets and the other having only one. $\endgroup$ – gorzardfu Jun 25 '16 at 15:19
  • $\begingroup$ (1) You said {1,3} is not open, but I thought in a discrete space all sets are both open and closed (2) I know the example is not really an isomorphism , but it seems like they fit the given definition. $\endgroup$ – Michael Maliszesky Jun 26 '16 at 3:47
  • $\begingroup$ @MichaelMaliszesky when we say a discrete space we mean a set endowed with the topology being the power set of that set. Neither of the topologies you have mentioned are the power set of X, so they are not discrete spaces. But I know how it can get confusing because yeah, X is a finite set which can also mean discrete. Either way, a discrete space is one where every subset of the underlying set is open. $\endgroup$ – gorzardfu Jun 26 '16 at 3:50
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    $\begingroup$ I think I'm finally getting this! If there is not a metric than "openness" is defined by the sets listed in the topology. Thank you so much! $\endgroup$ – Michael Maliszesky Jun 26 '16 at 4:22
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A homeomorphism needs to be a map from a space to another, not a map from the topology of one space to the topology of another.

That is, if $(X,\mathscr{T}_X)$ and $(Y,\mathscr{T}_Y)$ are topological spaces, a homeomorphism should be a pointwise map be a map $g\colon X\rightarrow Y$ satisfying the properties you listed above. You have used instead a map $g\colon \mathscr{T}_X\rightarrow \mathscr{T}_Y$. Note that the latter requires a topology of topology to begin talking about continuity.

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  • $\begingroup$ Can you give an example of a pointwise map. And I think I do not get the difference between a "SET", "TOPOLOGY" and a "SPACE". I was under impression that a topology of a set IS a space. $\endgroup$ – Michael Maliszesky Jun 25 '16 at 14:54
  • $\begingroup$ In your example, $X=\{1,2,3\}$ is the space and $\mathscr{T}_1$ and $\mathscr{T}_2$ are two possible topologies for it. $g\colon X\rightarrow X$ defined by $g(1)=2$, $g(2)=3$, and $g(3)=1$ is a pointwise map. $\endgroup$ – parsiad Jun 25 '16 at 14:55
  • $\begingroup$ I would recommend an introduction to simple (nonaxiomatic) set theory before you get into topology. Doing some real analysis (with metric spaces) before topology might be helpful too, since topology is a generalization of metric spaces. $\endgroup$ – parsiad Jun 25 '16 at 14:56
  • $\begingroup$ I did a under grad. class in real analysis already (culminating in proving the fundamental theorem of calculus), and when there are metrics involved things make more sense to me. In analysis we Dealt with sets, not spaces, and i'm not "getting" the difference. In this example, there isn't a metric, and I'm having trouble . $\endgroup$ – Michael Maliszesky Jun 25 '16 at 15:05
  • $\begingroup$ In real analysis, the space is $\mathbb{R}$, and the topology is defined by taking all open balls generated by the metric $d$ defined by $d(x,y)=|x-y|$. $\endgroup$ – parsiad Jun 25 '16 at 15:08

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