# How is $K \subset G_{\alpha_1} \cup \cdots \cup G_{\alpha_n}$, where $G_\alpha$'s are the open subsets of X corresponding to open subsets of $Y$?

Suppose $K\subset Y \subset X$. Then, $K$ is compact relative to X $\textit{iff}$ $K$ compact relative to Y.

Solution: The proof in the Rudin goes like this:

In this proof, I understand that there exist some $G_{\alpha}$ for each $V_{\alpha}$ but how to reach to the fact that $K$ is subset of finite collection of these $G_{\alpha}$'s and not other open sets in $X$ (or RHS of (22) is the open cover of $K$). Eq. (22) is what I don't understand.

Problem rephrase (Edit): For a set to be compact, we have to show that every open open cover of $K$ contains finite subcover. i.e. If ${G_{\alpha}}$ is an open cover of $K$ , then there are finitely many indices $\alpha_1,\cdots,\alpha_n$ such that $K \subset G_{\alpha_1} \cup \ldots \cup G_{\alpha_n}$. Now, we have proven that there are some open sets in $X$ correspoding to each $V_{\alpha}$. How do we make sure that these $G_{\alpha}$ form an open cover?

[This the from rudin's book section-2.33]

It's the definition of compactness relative to $X$ that whenever we have that $K$ is covered by a collection of open sets (relative to $X$), i.e. $K \subset \cup_\alpha G_\alpha$ (see def 2.31), then it is covered by finitely many of them, so as per 2.32 we have finitely many indices $\alpha_1,\ldots,\alpha_n$ such that $K \subset G_{\alpha_1} \cup \ldots \cup G_{\alpha_n}$.

So that step is just applying the definitions that were just given. It's the same formula as in 2.32.

Because we want to show that $K$ is compact relative to $Y$ we start with an open cover of $K$ by open sets relative to $Y$ (and we need a finite subcover of those open sets); these are the $V_\alpha$. Because the assumption on $K$ is compactness relative to $X$ we need to go from the $V_\alpha$ to the corresponding $G_\alpha$ (which are open relative to $X$). Because the open sets are related by $G_\alpha \cap Y = V_\alpha$ we can use the finite cover for the corresponding cover to go back to the original cover $V_\alpha$. If we'd use other open sets we would lose this connection.

We know that $K$ is covered by the finitely many $V_{\alpha_i}$. So if $x \in K$, we know that $x \in V_{\alpha_{i(x)}}$ for some $i(x) \in \{1,\ldots,n\}$. We know by assumption that $x \in Y$, as $K \subset Y$. So $x \in V_{\alpha_{i(x)}} \cap Y = G_{\alpha_{i(x)}}$. As this works for any $x$, $K$ is covered by the $G_{\alpha_1},\ldots,G_{\alpha_n}$ for the same indices that work for the $V_\alpha$.

• That I understand but look at $G_{\alpha}$, these are obtained from $V_{\alpha}$ and I don't understand why $G_{\alpha}$ should be the open cover and not any other open sets in X. It's not the finiteness but the specific set of $G_{\alpha}$ that's bothering me. – MUH Jul 12 '16 at 13:15
• @MUH added an extra explanation, which I hope helps. – Henno Brandsma Jul 12 '16 at 15:00
• We have to prove that this condition is necessity (not that we are doing it to maintain the required connection), only then we will be able to prove that $K$ is relatively open in $Y$. I am rephrasing the question again so that you have a better idea of my doubt. – MUH Jul 13 '16 at 10:40
• For a set to be compact, we have to show that every open open cover of $K$ contains finite subcover. i.e. If ${G_{\alpha}}$ is an open cover of $K$ , then there are finitely many indices $\alpha_1,\cdots,\alpha_n$ such that $K \subset G_{\alpha_1} \cup \ldots \cup G_{\alpha_n}$. Now, we have proven that there are some open sets in $X$ correspoding to each $V_{\alpha}$. How do we make sure that these $G_{\alpha}$ form an open cover? – MUH Jul 13 '16 at 10:46
• @MUH added some more explanation why they cover. – Henno Brandsma Jul 15 '16 at 15:33