A topological space is compact iff each infinite subset has a complete accumulation point

Question

This is based on the comments and answers provided in this post. However, I have some questions on the proof and the hint given in Kelleys book p.163. I will highlight the hint of the book. My own thoughts will be preceded by a dot.

If $X$ is not compact choose an open cover $A$ with no finite subcover such that the cardinal number $c$ of $A$ is as small as possible.

• This is possible since assuming $X$ not to be compact there is at least one open cover $\mathcal{O}$ without finite subcover.
• If one then considers the non empty set $\mathfrak{M}$ of all coverings without finite subcovers, the well-ordering principle implies the existence of $A$ as above, i.e. an element of $\mathfrak{M}$ with minimal cardinality.

Let $C$ be a well-ordered set of cardinal $c$ such that the set of predecessors of each member has a cardinal less than $c$ (It is shown in the appendix that $c$ is such a set.)

• If we take the ordinal number corresponding to the cardinality $c$ this should be satisfied.

Let $f$ be a one-to-one map of $C$ onto $A$. Then for each member $b$ of $C$ the union $$\left[U_b :=\right]\quad \bigcup \, \{f(a):a<b\}$$ does not cover $X$

• because if it did, there would exist an cover with cardinality strictly less than $c$.

and, in fact, the complement of this union must have cardinal number at least as great as $c$.

• Why?

It is therefore possible to choose $x_b$ from the complement such that $x_a \neq x_b$ for $a < b$.

• Why?

Consider the set of all $x_b$.

• Denote this set with $M$. It is infinite because for each Union $U_b$ there exists $x_b \in M$ with $x_a \neq x_b$ for $a < b$. This implies $|M| = c$.
• To show that $M$ has no limit point, consider $x \in X$. The covering property gives $O_x \in A$ such that $x \in O_x$. Then the following calculation should hold $$ |M \cap O_x | \leq | \{x_b \mid b \leq a \} | \leq | a | < c = |M|.$$
• Consequently $M$ has no limit point.

Did I make any mistakes so far? Any suggestions on the points of the proof that still remain unclear to me?

Answer 1

For $b \in C$ let $A_b := \{ f(a) : a < b \}$ (so that $U_b = \bigcup A_b$).

• I'll back up a bit, and give a fairly full justification for why $A_b$ cannot cover $X$.

  If $A_b$ covers $X$ (so that $U_b = X$), since $A_b$ is a subfamily of $A$ it must be that $A_b$ does not have a finite subcover (otherwise $A$ would). But the choice of the well-ordered set $C$ implies that $| A_b | < c$, which contradicts the choice of the cardinal $c$! Therefore $A_b$ cannot cover $X$.

• If $| X \setminus U_b | < c$, then for each $x \in X \setminus U_b$ pick any $V_x \in A$ containing $x$. Then $A_b \cup \{ V_x : x \in X \setminus U_b \}$ covers $X$ and is a subfamily of $A$. Similarly to the above, this cannot have a finite subcover, and it has cardinality $< c$, which then contradicts our choice of $c$. Therefore $| X \setminus U_b | \geq c$.

• Given any $b \in C$, as $\{ x_a : a < b \}$ has cardinality $< c$ and as $| X \setminus U_b | \geq c$, it follows that $X \setminus ( U_b \cup \{ x_a : a < b \} )$ is nonempty, and so we may choose $x_b$ from this set arbitrarily to be distinct from $x_a$ for $a < b$.

• Note that the set $M = \{ x_b : b \in C \}$ has cardinality $c$. It has no complete accumulation point because for each $x \in X$ if $b \in C$ is least such that $x \in U_b$, then $U_b$ is an open neighbourhood of $x$ and $U_b \cap M \subseteq \{ x_a : a < b \}$, and so has cardinality $<c$.