So Im aware this question has been asked already here.
However, the definition of compactness for a subset I am following is slightly different and I am struggling to translate the top answer for the question to suit my definition.
The definition I am following is:
Let $X$ be a topological space and let $S\subseteq X$. Then $S$ is a compact set if $S$ with the subspace topology is a compact topological space.
The problem is that this definition means that if $\mathscr{U}$ is a cover of $S_1\cup S_2$ then $S_1\cup S_2=\bigcup\mathscr{U}$ as appose to $S_1\cup S_2\subseteq\bigcup\mathscr{U}$ and so I don't know how deduce that $\mathscr{U}$ is open cover for $S_1$ and for $S_2$
Best Answer
Let $\mathcal U$ be a (relative, in the subspace topology of $S$) open cover of $S=S_1\cup S_2$. For each $U\in\mathcal U$ fix $U'$ open in $X$ with $U=U'\cap S$. Let $\mathcal U_1=\{U'\cap S_1:U\in\mathcal U\}$ and $\mathcal U_2=\{U'\cap S_2:U\in\mathcal U\}$. Then $\mathcal U_1$ is a (relative) open cover of $S_1$ and hence has a finite subcover $\{V_1,\dots,V_n\}$. Similarly, $\mathcal U_2$ is a (relative) open cover of $S_2$ and hence has a finite subcover $\{W_1,\dots,W_m\}$. For each $j=1,\dots,n$ fix $U(1,j)\in\mathcal U$ with $V_j=U'(1,j)\cap S_1$. For each $k=1,\dots,m$ fix $U(2,k)\in\mathcal U$ with $W_k=U'(2,k)\cap S_2$. Verify that $\{U(1,j):j=1,\dots,n\}$ together with $\{U(2,k):k=1,\dots,m\}$ is a finite (relatively) open cover of $S$ (and it is a subcover of $\mathcal U$).