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Every metric space which is both $T_1$-space and normal space is Hausdorff space.

$(X,\tau)$ is Hausdorff if $\forall x,y\in X\exists U,V\in\tau$ such that $x\in U$ and $y\in V$ and $U\cap V=\emptyset$.

It easy to note that every Hasudorff space is $T_1$-space. However I know that every metrizable space is Hausdorff therefore it is $T_1$-space. However I cannot necessarily see or coneceive a proof that a metric space which is both $T_1$-space and normal space is Hausdorff space.

Question:

Can anyone help me provide a proof?

Thanks in advance!

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Hint: In a $T_1$ space points are closed.

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If a topological space is normal, then given any two disjoint closed sets $F_1,F_2$ there exist disjoint open sets $U_1,U_2$ such that $F_1\subset U_1$ and $F_2\subset U_2$.

If a topological space is $T_1$, then any singleton is closed. Therefore, if a topological space is $T_1$ and normal, then given any two distinct points $p_1,p_2$, the sets $\{p_1\}$ and $\{p_2\}$ are disjoint closed sets, so by normality, there exist disjoint open sets $U_1,U_2$ such that $p_1\in U_1$ and $p_2\in U_2$, which means that the space is Hausdorff.

There is no need to assume that the space has a metric. In fact, in a metric space, all finite subsets are closed, so in particular, every metric space is $T_1$. Also, every metric space is normal, so every metric space is also Hausdorff. Heuristically speaking, if you can measure distance between distinct elements, then you can separate them.

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