# Prove that $(X,\tau_1\cap\tau_2)$ is also a $T_1$-space, whenever $\tau_1$ and $\tau_2$ are $T_1$.

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.

• If $\{x\}\in \tau_1$ and $\{x\}\in \tau_2$ then $\{x\}\in \tau_1\cap \tau_2$ by the definition of intersection. Nov 25, 2022 at 13:40
• But you should consider the complements of one-point-sets.
– Ulli
Nov 25, 2022 at 16:06
• @Ulli I tried using the complements the first time I attempted the problem but the instructor said she wants us to prove the singleton sets are closed. Why do you think she would prefer that statement over the complements which are open in $\tau_1$ and $\tau_2$? Nov 27, 2022 at 3:17
• @cantor's sloth: Singletons are closed, iff their complements are open (which holds by definition for every subset). If your instructor does not allow you to use this, you should really ask her for the solution.
– Ulli
Nov 27, 2022 at 22:14

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 $$T_1$$ $$\square$$.
• What is that fancy $C\{x\}$? I've never seen that. Nov 27, 2022 at 3:19
• Given a set $U$ and a set $A\subseteq U$, we define $\mathscr{C}_U A=U\setminus A$ and is called the complement of $A$ with respect to $U$. When unambiguous, the subindex can be omitted. The LaTeX code for that font is "\mathscr": You can use right click on any math in the site to see the LaTeX code for anything. Nov 27, 2022 at 4:26