Excluded Point Topology is Topology

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Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.

Then $\tau_{\bar p}$ is a topology on $S$, and $T$ is a topological space.


Proof

We have by definition that $S \in \tau_{\bar p}$, and as $p \notin \varnothing$ we have that $\varnothing \in \tau_{\bar p}$.


Now let $U_1, U_2 \in \tau_{\bar p}$.

By definition $p \notin U_1$ and $p \notin U_2$, and so $p \notin U_1 \cap U_2$ by definition of set intersection.

So $U_1 \cap U_2 \in \tau_{\bar p}$.


Now let $\mathcal U \subseteq \tau_{\bar p}$.

We have that $\forall U \in \mathcal U: p \notin U$.

Hence from Subset of Union $p \notin \bigcup \mathcal U$.


So all the properties are fulfilled for $\tau_{\bar p}$ to be a topology on $S$.

$\blacksquare$


Sources