Point in Finite Hausdorff Space is Isolated/Proof 1

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

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

Let $x \in X$.

Then $x$ is isolated in $X$.


As $X$ is finite, its elements can be placed in one-to-one correspondence with the elements of $\left\{{1, 2, \ldots, n}\right\}$ for some $n \in \N$.

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

From the definition of Hausdorff space:

$\forall x_i, x_j \in X: x_i \ne x_j: \exists U, V \in \tau: x_i \in U, x_j \in V: U \cap V = \varnothing$

Let $x_k \in X$.

Then by definition:

$\forall x_i \in X, x_i \ne x_k: \exists U_i, V \in \tau: x_k \in U_i, x_i \in V: U_i \cap V = \varnothing$

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

That is:

$\forall x_i \in X: \exists U_i \in \tau: x_k \in U_i, x_i \notin U_i$

Now consider:

$\displaystyle U = \bigcap_{i \mathop \ne k} U_i$

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

We have that:

$\forall i: x_k \in U_i$
$\forall i: x_i \notin U_i$


$U \cap X = \left\{{x_k}\right\}$

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.