Let $\tau_1$ and $\tau_2$ be two topologies on a set $X$ and that $(X,\tau_1)$ and $(X,\tau_2)$ are $T_1$-spaces (discrete topology). Prove that $(X,\tau_3)$ is also a $T_1$-space, where $\tau_3=\tau_1\cap\tau_2$.
Definition: A topological space $(X,\tau)$ is said to be a $T_1$-space if every singleton set $\{x\}$ is closed in $(X,\tau)$.
Here's my proof:
Let $\tau_1$ and $\tau_2$ be two topologies on a set $X$ such that $(X,\tau_1)$ and $(X,\tau_2)$ are $T_1$-spaces. By definition of a $T_1$-space, for all $x\in X$ the singleton sets $\{x\}\in X$ are closed both in $\tau_1$ and $\tau_2$. Since $\tau_3$ was already shown to be a topology on $X$ in the previous problem, the intersection of $\tau_1$ and $\tau_2$ is also in $\tau_3$. Thus all the singleton sets are closed in $\tau_3$ making $\tau_3$ a $T_1$-space as desired. $\square$
Please help as I know this isn't correct but I'm stuck and not sure what to do to prove this. Thanks in advance.
Best Answer
First of all, it's worth mentioning that $\tau_3$ is a topology since it's the intersection of topologies.
Let $x\in X$. Since $\tau_1$ and $\tau_2$ are $T_1$, $\{x\}$ is closed in both of them. That is, $\mathscr{C}\{x\}\in\tau_1\cap\tau_2$. From which follows that $\mathscr{C}\{x\}\in\tau_3$. We conclude that $\{x\}$ is closed in $(X,\tau_3)$ and thus it is $T_1$ $\square$.