Show that $T_1$ axiom is equivalent to the condition that for each pair of points of $X$, each has a neighbourhood not containing the other.
I rephrase this exercise to my taste.
Topological space $(X,\mathcal{T}_X)$ satisfy $T_1$ axiom $\iff$ $\forall x_1,x_2\in X,x_1 \neq x_2$ there exists an open set $U_1$ such that $x_1 \in U_1$ but $x_2 \not\in U_1$, and there exists an open set $U_2$ such that $x_2 \in U_2$ but $x_1 \not\in U_2$.
My attempt: ($\Rightarrow$) Suppose finite sets are closed in $X$. Let $x,y \in X$ such that $x\neq y$. Since $\{x\}$ and $\{y\}$ is closed in $X$, we have $X-\{x\} \in \mathcal{T}_X$ and $X-\{y\}\in \mathcal{T}_X$. So $\exists X-\{x\}\in \mathcal{T}_X$ such that $y \in X- \{x\}$ and $x\notin X-\{x\}$. $\exists X-\{y\}\in \mathcal{T}_X$ such that $x \in X- \{y\}$ and $y\notin X-\{y\}$.
($\Leftarrow$) It is suffices to show singleton sets are closed. fix $x\in X$. Let $y\in X$ such that $y\neq x$. $\exists U_y\in \mathcal{T}_{X}$ such that $y\in U_y$ and $x\notin U_y$. So $U_y \cap \{x\} = \phi$. By theorem 17.5, $y\notin \overline{ \{x\} }$(Note $U_y$is the neighbourhood of y). Thus $\overline{ \{x\} }= \{x\}$, $\{x\}$ is closed. Is this proof correct?
Best Answer
The proof is in essence correct. Maybe some notational improvement and added or alternate reasoning wouldn't hurt, e.g. in my edited version:
Suppose finite sets are closed. Let $x_1,x_2 \in X$ be such that $x_1 \neq x_2$. Then $U_2:= X-\{x_1\}$ is open as $\{x_1\}$ is closed, and likewise $U_1:=X-\{x_2\}$ is open. As $x_1 \notin U_2,x_2 \in U_2$ and $x_2 \in U_2, x_1 \notin U_2$, your reformulated version is satisfied indeed. (using the same variable names is clearer IMO)
It suffices to show that singleton sets are closed (being closed is preserved by finite unions and a finite set is a finite union of singletons). Suppose that $\{x\}$ is not closed for some $x \in X$. This means that for some $y \neq x$ we have $y \in \overline{\{x\}}$. By 17.5 this means that every neighbourhood of $y$ intersects $x$ and this directly contradicts the $T_1$ reformulation for the pair $x,y$. So $\{x\}$ is closed and we're done.