Question about Definition of homeomorphism (counter example) 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?

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 $Y$.
