Recently in our UCLA summer analysis course we discussed the equivalence of compactness and sequential compactness for subsets of a metric space. I thought of a proof which I thought to be considerably shorter and simpler than those usually given. However it requires Zorn's Lemma, so it is perhaps not as fundamental.
Theorem. A sequentially compact subset of a metric space is compact.
Proof. Suppose $A \subset M$ (where $M$ is a metric space) has an open cover with no finite subcover. Let this cover be given by $C=\{O_i\}_{i\in I}$, where $I$ is an arbitrary indexing set (could be uncountable). Subcovers of $C$ are partially ordered by inclusion, and every chain of subcovers has a minimal element (given by the intersection of all of its elements). Thus by Zorn's Lemma we can assume that ${C}$ has a minimal subcover, which we will call ${C}'=\{O_k\}_{k\in K}$.
We now argue that each $O_k \in C'$ has an element $x_k$ such that $x_k \in {O}_j$ if and only if $j=k$. Indeed if we had an open set $O_k$ such that every element $x \in O_k$ was contained in some other open set in $C'$, the collection ${C}''={C}'-\{O_k\}$ would be an open cover and obviously a proper subset of ${C}'$, contradicting the assumption that ${C}'$ is a minimal subcover.
By the axiom of choice ${C}'$ has a countable subset $\{O_1,\ldots \}$. Out of each $O_i$ we pick an element $o_i$ such that $o_i \in O_j$ if and only if $i=j$ (guaranteed to exist by previous paragraph). Thus $\{o_i\}_{i=1}^\infty$ is a sequence. It could not have a convergent subsequence because then the limit point $o$ would be in some set $O_m \in {C}'$, which would then necessarily contain some $o_i$ with $i \ne m$.
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This comment will most likely not be noticed, because this blog does not seem to be very active.
However, I must (as a student of mathematics at Helsinki University) proove that this "proof" is inaccurate. This proof assumes metricability, but does not use it. Thus this proof, if ture, would imply that for ANY Topological space X sequential compactness would imply compactness.
For any informed reader this is an impossible implication. Thus the proof contains an error. This error is in the very first paragraph. Namely the intersection of a family o covers, which form a chain, is NOT NECESSARILY a cover anymore. If we take a finite intersection, this remains true. Let me show an example:
D is the unit disc centered at the origin. For every point x in D different from the origin choose a neighbourhood such that it does not contain the origin, B_x let us call them. The for the origin choose the following sets: B(0,r) for all R \> r \> 0, where B(0,r) are the balls centered at the origin with radius r.
The families we now examine are F_R={B_x : x in D} union {B(0,r) : 0 \< r \< R}. Now each F_R is a cover, but the intersection of F_R:s for R \> 0 does not contain ANY neighbourhood for the origin and thus the origin is not contained in them. So the intersection of F_R is NOT anymore a cover of D.
So by this counterexample I have shown that the utilization of Zorns lemma in this manner is inaproppriate.
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