Topology. we have to prove $\mathbb{X}$ is hausdorff space on real valued continuous function.

Question

If, for any two distinct points $x$ and $y$ in a topological space $\mathbb{X}$, there is always a real-valued continuous function such that $f(x) ≠ f(y)$, prove that $\mathbb{X}$ is Hausdorff.

Answer 1

Since $f(x)\neq f(y)$, you can create two open sets $U,V\subset \mathbb{R}$ such that $f(x)\in U$, $f(y)\in V$ and $U\cap V=\emptyset$.

For example, $U=\{r\in\mathbb{R}:|r-f(x)|<\frac{|f(x)-f(y)|}{2}\}$ and $V=\{r\in\mathbb{R}:|r-f(y)|<\frac{|f(x)-f(y)|}{2}\}$. These are two open intervals that don't intersect.

Now, $f^{-1}(U)$ and $f^{-1}(V)$ don't intersect either. (For example, every $z\in f^{-1}(U)$ has the property that $f(z)\in U$. But if $f(z)\in U$, then $f(z)\notin V$. Remember, $U$ and $V$ don't intersect.)

But $f^{-1}(U)$ and $f^{-1}(V)$ are also open sets.

This proves the Hausdorff property.