# In proving “if a set is compact, then it must be closed”, why does the finite subcover behave differently than the infinite open cover?

'''If a set is compact, then it must be closed.'''

Let $$S$$ be a subset of $$\mathbb{R}^n$$. Observe first the following: if $$a$$ is a limit point of $$S$$, then any finite collection $$C$$ of open sets, such that each open set $$U \in C$$ is disjoint from some neighborhood $$V_U$$ of $$a$$, fails to be a cover of $$S$$. Indeed, the intersection of the finite family of sets $$V_U$$ is a neighborhood $$W$$ of $$a$$ in $$\mathbb{R}^n$$. Since $$a$$ is a limit point of $$S$$, $$W$$ must contain a point $$x$$ in $$S$$. This $$x$$$$S$$ is not covered by the family $$C$$, because every $$U$$ in $$C$$ is disjoint from $$V_U$$ and hence disjoint from $$W$$, which contains $$x$$.

If $$S$$ is compact but not closed, then it has an accumulation point $$a$$ not in $$S$$. Consider a collection $$C’$$ consisting of an open neighborhood $$N(x)$$ for each $$x \in S$$, chosen small enough to not intersect some neighborhood $$V_x$$ of $$a$$. Then $$C’$$ is an open cover of $$S$$, but any finite subcollection of $$C’$$ has the form of $$C$$ discussed previously, and thus cannot be an open subcover of $$S$$. This contradicts the compactness of $$S$$. Hence, every accumulation point of $$S$$ is in $$S$$, so $$S$$ is closed.

Why must the set $$C$$ in the first paragraph be finite? Shouldn't it be true for all $$C$$ whether or not it's finite?

And in the second paragraph, why is $$C’$$ an open cover but any finite subcollection of $$C’$$ is not an open subcover? There must be some fundamental difference I’m missing in my current thinking: If $$C’$$ always excludes some neighborhood $$V_x$$ of $$a$$ for all $$x$$, then even the infinite collection of $$C’$$ should not be an open cover!

Take the interval $(0,1]$, and the infinite cover consisting of the intervals $(\frac1n,1]$. The whole interval is covered, but no finite subset covers the interval. Since I've just given you an open cover with no finite subcover, $(0,1]$ is not compact.

In this context, $a=0$ and $C' = \{(\frac1n,1]\mid n\in \Bbb Z\}$ (note how $a$ is not an element of $S$). I hope it illustrates how an infinite cover where each open set misses a neighbourhood of an accumulation point, still can cover the whole set.

In the first paragraph, $C$ must be finite since what we've proven is that the union of the sets in $C$ is disjoint from the intersection of the neighbourhoods $V_U$. This is only guaranteed to be open if it is a finite intersection.

For infinite $C$ the claim need not be true because the intersection of infinitely many open $V_u$ need not be open. For example $\bigcup_{n\in\mathbb N}\left]-\frac1n,\frac1n\right [$ equals $\{0\}$, which is not open. Same applies for the other question: An intersection of infinitely many open $V_x$ need not be open. $C'$ is an open cover because for each $x$ we have $x\in N(x)\in C'$.

The collection $C$ must be finite in order to ensure that $\bigcap_{U\in C}V_U$ is open: the intersection of a finite family of open sets is always open, but the intersection of an infinite family of open sets need not be open. In the second paragraph $C'=\{N(x):x\in S\}$, where $N(x)$ is an open nbhd of $x$ for each $x\in S$. Thus, for each $x\in S$ we have $x\in N(x)\subseteq\bigcup C'$, and therefore $S\subseteq\bigcup C'$, i.e., $C'$ is an open cover of $S$.

To answer your first question: No, for example, consider $S=(0,1)$, the limit point $0$, and the open cover of $S$ consisting of the sets $(1/n,1)$ for $n\geq 1$. Each set $(1/n,1)$ is disjoint from a neighborhood of $0$, for example $(-1,1/2n)$.

To answer your second question: $C'$ is an open cover by construction. For each $x\in S$, $C'$ contains a neighborhood $N(x)$, so it automatically covers all of $S$. Note $C'$ does not need to cover $a$ because $a$ is assumed not to be in $S$.