Point in Finite Hausdorff Space is Isolated/Proof 1

Theorem
Let $T = \left({S, \vartheta}\right)$ be a Hausdorff space.

Let $X \subseteq S$ such that $X$ is finite.

Let $x \in X$.

Then $x$ is isolated in $X$.

Proof
As $X$ is finite, its elements can be placed in one-to-one correspondence with the elements of $\N^*_n$ for some $n \in \N$.

So let $X = \left\{{x_1, x_2, \ldots, x_n}\right\}$.

From the definition of Hausdorff space:
 * $\forall x_i, x_j \in X: x_i \ne x_j: \exists U, V \in \vartheta: x_i \in U, x_j \in V: U \cap V = \varnothing$

Let $x_k \in X$.

Then by definition:
 * $\forall x_i \in X, x_i \ne x_k: \exists U_i, V \in \vartheta: x_k \in U_i, x_i \in V: U_i \cap V = \varnothing$

Note that $U_i$ may be different for each $X_i$.

That is:
 * $\forall x_i \in X: \exists U_i \in \vartheta: x_k \in U_i, x_i \notin U_i$

Now consider:
 * $\displaystyle U = \bigcap_{i \ne k} U_i$

We have by definition of a topology that $U \in \vartheta$, as the intersection $U$ is of a finite number of sets $U_i$.

We have that:
 * $\forall i: x_k \in U_i$
 * $\forall i: x_i \notin U_i$

So:
 * $U \cap X = \left\{{x_k}\right\}$

and so $U$ is an open set of $T$ which contains no points of $X$ other than $x_k$.

So by definition, $x_k$ is an isolated point of $X$.

Hence the result.