What does it mean when two sets are "adjoined" in a metric space?

I encountered the word "adjoined" in Baby Rudin, Chapter 2 concerning basic topology on Euclidean space. It appeared in the proof to Theorem 2.35

Theorem$\quad$ Closed subsets of compact sets are compact
Proof$\quad$ Suppose $F\subset K\subset X$, $F$ is closed (relative to $X$), and $K$ is compact. Let $\{V_{\alpha}\}$ be an open cover of $F$. If $F^c$ is adjoined to $\{V_{\alpha}\}$ ,we obtain an open cover $\Omega$ of $K$. Since $K$ is compact, there is a finite subcollection $\Phi$ of $\Omega$ which covers $K$, and hence $F$. If $F^c$ is a member of $\Phi$, we may remove it from $\Phi$ and still retain an open cover of $F$. We have thus shown that a finite subcollection of $\{V_{\alpha}\}$ covers $F$.

First, to be honest I don't understand the meaning of "adjoined", I guess it might mean that the two sets are "complementary" except for the "boundary" between them. If this is what "adjoined" means, then I am still confused. Because since $\{V_{\alpha}\}$ is an open cover of $F$, and $F^c$ is open. If these two open sets are "adjoined", then neither of them includes $\partial F$, which is absurd. So it might be that I just failed to understand the word "adjoined" properly. Without a proper understanding, I find it hard for me to process the whole proof.

Can you help me? Thanks in advance!

Solution 1:

The adjoining simply means taking a union of $F^C$ and $\\{V_{\alpha}\\}$.

Since $F$ is closed, $F^C$ is open. Also we assumed $K$ is compact, hence, $K$ has a finite open cover, say $\\{V_{\alpha}\\}$ which covers $K$. Adjoining $F^C$, or more simply put, taking union of $F^C$ and $\\{V_{\alpha}\\}$ will also cover $K$. Let us call $F^C\cap V_{\alpha}=\Omega$. At this point, it is easy to see that $\Omega$ is a finite collection and covers $K$ and therefore, also $F$. We can choose a subset $\Phi$ of this finite collection $\Omega$ that would be sufficient to cover $F$ and hence, we proved that $F$ is compact.