Let
- $(E,\tau)$ be a topological space and $\Delta\not\in E$;
- $E_\Delta:=E\cup\{\Delta\}$ and $$\tau_\Delta:=\tau\cup\underbrace{\{E_\Delta\setminus B:B\subseteq E\text{ is closed and compact}\}}_{=:\:\sigma_\Delta}.$$
We can easily show that $\tau_\Delta$ is a topology on $E_\Delta$.
How do we show that $\tau_\Delta$ is compact and why is it necessary to integrate the "closedness" in the definition of $\sigma_\Delta$?
Let $\mathcal B_\Delta\subseteq2^{E_\Delta}$ be an $\tau_\Delta$-open cover of $E_\Delta$.
Obviously, there must be a $B_\Delta\in\mathcal B_\Delta$ with $\{\Delta\}\subseteq B_\Delta$ and, by construction of $\tau_\Delta$, $$B_\Delta=E_\Delta\setminus B\tag1$$ for some $\tau$-closed $\tau$-compact $B\subseteq E$.
Now $\mathcal B_\Delta$ is clearly an $\tau_\Delta$-open cover of $B$ as well. So, there is a $\mathcal B\subseteq\mathcal B_\Delta$ with $$B\subseteq\bigcup\mathcal B\tag2.$$
However, why can we assume that $\mathcal B\subseteq\tau$?
Assuming $\mathcal B\subseteq\tau$, since $B$ is $\tau$-compact, there is a finite $\mathcal F\subseteq\mathcal B$.
How do we now conclude that $\mathcal F\cup\{B_\Delta\}$ is a finite subcover of $\mathcal B_\Delta$? And where did we need the closedness?
We want that $E_\Delta$ contains $E$ (with its original topology!) as a subspace. This requires that all $(E_\Delta \setminus B) \cap E = E \setminus B$ belong to $\tau$, i.e. that all $B$ are closed in $E$.
We cannot. But we know that $B\subset \bigcup\mathcal B$. Since $B$ is a compact subspace of $E$ and $E$ is a subspace of $E_\Delta$, we see that $B$ is a compact subspace of $E_\Delta$. Therefore there exist finitely many $U_i \in \mathcal B$ such that $B \subset \bigcup U_i$. This completes the proof.