# If $X$ is a set and $\tau_1$ is finer than $\tau_2$, prove if $(X, \tau_2)$ is Hausdorff, then $(X, \tau_1)$ is Hausdorff.

I tried to do this by contradiction.

So we have that $(X, \tau_2)$ is Hausdorff, and $\tau_2 \subset \tau_1$.

Suppose that $(X, \tau_1)$ was not Hausdorff. Then we have elements $y,z \in X$ where $z\neq y$ and two open sets, call them $U$ and $W$ where $y \in U$, $z \in W$ and $U \cap W \neq \emptyset$.

I'm kinda stuck here, I wanna use that $\tau_1$ is finer than $\tau_2$ somehow, but I can't really figure it out.

Any help would be great!

• What is the definition of finer? – graydad Jan 28 '15 at 23:43
• @graydad Actually... that's a great question. I should have looked up that definition before asking the question, which makes my question seem a bit stupid now... I'll edit and provide an answer. – Jan Jan 28 '15 at 23:45
• when you see the definition, you should have no trouble getting it :) – graydad Jan 28 '15 at 23:46
• @Jan You don't know the definition? Then were did $\tau_2\subset \tau_1$ come from? – Git Gud Jan 28 '15 at 23:46
• Contradiction not needed, we need to show that for any $x$, $y$ with $x\ne y$, there are $U$ and $V$ in in $\tau_1$ such that $\dots$. There are such $U$ and $V$ in $\tau_2$, and by "finer" these are in $\tau_1$. – André Nicolas Jan 29 '15 at 0:25

if one topology is finer then for $x,y\in X$ you can take the same open sets that are disjoint neighbourhoods of $x$ and $y$ in both topologies.
• Ah, from what I have... this contradictions my assumption that $(X, \tau_2)$ is Hausdorff because if $(X, \tau_1)$ is not Hausdorff, then it's impossible for $(X, \tau_2)$ to be Hausdorff as well. – Jan Jan 28 '15 at 23:49