if a topology $\tau$ is finer than the usual topology $\tau_0$ on $\mathbb{R}$ then ($\mathbb{R}$,$\tau$ ) is Hausdorff

Question

A topology $\tau$ is finer than $\tau_0$ if $\tau_0 \subset \tau$. I know that ($\mathbb{R}$,$\tau_0$) is Hausdorff where $\tau_0$ is the usual topology on $\mathbb{R}$. Now, why ($\mathbb{R}$,$\tau$ ) is also Hausdorff?

Answer 1

To show that $(\mathbb{R},\tau)$ is Hausdorff, we need to prove that for any pair of distinct points $x\neq y$ in $\mathbb{R},$ there exist $U\in\tau$ and $V\in\tau$ of $x$ and $y$ respectively such that $U\cap V=\emptyset.$

Fix $x\in\mathbb{R}$ and $y\in\mathbb{R}$ such that $x\neq y.$ Since $(\mathbb{R},\tau_0)$ is Hausdorff, there exist $U\in \tau_0$ and $V\in \tau_0$ such that $x\in U, y\in V$ and $$U\cap V=\emptyset. $$ As $\tau_0\subseteq \tau,$ so $U\in\tau$ and $V\in\tau.$ Therefore, $(\mathbb{R},\tau)$ is Hausdorff.

Answer 2

Let, $x,y\in(\mathbb{R},\tau)$ s.t. $x\neq y$. Since, $\tau_0\subset \tau$, so $\exists$ open sets $U,V\in \tau_0\subset\tau$ s.t. $x\in U$ $y\in V$ with $U\cap V= \emptyset$ (since $(\mathbb{R},\tau_0)$ is Hausdroff). Hence $(\mathbb{R},\tau)$ is Hausdroff.