# Prob. 4, Sec. 28 in Munkres' TOPOLOGY, 2nd ed: For $T_1$-spaces countable compactness is equivalent to limit-point-compactness.

Definition: A topological space $$X$$ is said to be countably compact if every countable open covering of $$X$$ has a finite subcollection that also covers $$X$$.

Definition: A topological space $$X$$ is said to be limit-point-compact if every infinite subset of $$X$$ has a limit point in $$X$$.

Then how to prove the following result?

Let $$X$$ be a $$T_1$$-space. Then $$X$$ is countably compact if and only if $$X$$ is limit point compact.

My effort:

Suppose $$X$$ is countably compact. If $$X$$ is not limit point compact, then let $$A$$ be an infinite subset of $$X$$ such that $$A$$ has no limit point in $$X$$.

Let $$B \colon= \left\{ b_1, b_2, b_3, \ldots \right\}$$ be a countably infinite subset of $$A$$.

Since we have assumed that $$A$$ has no limit point in $$X$$ and since $$B \subset A$$, therefore $$B$$ has no limit point in $$X$$ either. So, the set $$B^\prime$$ of all the limit points of $$B$$ in $$X$$ is empty and thus contained in $$B$$; hence $$B$$ is closed in $$X$$.

Since $$B$$ has no limit points in $$X$$ and since $$B \subset X$$, none of the elements of the set $$B$$ itself is a limit point of $$B$$; so for each element $$b_n \in B$$, there is an open set $$U_n$$ in $$X$$ such that $$U_n \cap B = \left\{b_n \right\} \tag{1} .$$

Now the collection $$\left\{ \ U_n \ \colon \ n \in \mathbb{N} \ \right\} \bigcup \{ X-B \}$$ forms a countable open covering of the countably compact space $$X$$, so some finite subcollection of this covering also covers $$X$$ and hence that finite subcollection also covers $$B$$. But the set $$X-B$$ contains no point of $$B$$. So $$B$$ is covered by finitely many of the sets $$\left\{ \ U_n \ \colon \ n \in \mathbb{N} \ \right\},$$ each of which contains exactly one point of set $$B$$ [Refer to (1) above.]. This implies that set $$B$$ is a finite set, contrary to our choice of $$B$$. Hence $$X$$ is limit point compact.

Is this proof correct?

How to prove the converse?

PS:

From the above proof (where we haven't required $$X$$ to be a T$$_1$$-space), we can even state the following:

Every (countably) compact topological space $$X$$ is also limit point compact.

Am I right?

P.S.:

Based on the answers below, I state and prove the following result:

Let $$X$$ be a $$T_1$$ topological space. If $$X$$ is limit point compact, then $$X$$ is also countably compact.

Proof:

Suppose that $$X$$ is a limit point compact, $$T_1$$ topological space that is not countably compact. Then there is a countable open covering $$\left\{ \ U_n \ \colon \ n \in \mathbb{N} \ \right\}$$ of $$X$$ that has no finite subcollection that also covers $$X$$.

Let us define the collection $$\left\{ \ V_n \ \colon \ n \in \mathbb{N} \ \right\}$$ of sets as follows: $$V_n \colon= \begin{cases} U_1 \ & \mbox{ if } \ n = 1, \\ V_{n-1} \cup U_n \ & \mbox{ if } \ n = 2, 3, 4, \ldots. \end{cases}$$ That is, \begin{align} V_1 & \colon= U_1, \\ V_2 & \colon= U_1 \cup U_2, \\ V_3 & \colon= U_1 \cup U_2 \cup U_3, \\ V_4 & \colon= U_1 \cup U_2 \cup U_3 \cup U_4, \\ & \cdots \\ \end{align}

We note that $$V_1 \subset V_2 \subset V_3 \subset \cdots. \tag{1}$$

For each $$n \in \mathbb{N}$$, as $$U_n \subset V_n \subset X$$ and as $$\bigcup_{n=1}^\infty U_n = X, \tag{2}$$ so we must also have $$\bigcup_{n = 1}^\infty V_n = X. \tag{2*}$$ Thus $$\left\{ \ V_n \ \colon \ n \in \mathbb{N} \ \right\}$$ is also a countable open covering of $$X$$.

Now as $$V_1 = U_1$$ is a proper subset of $$X$$ [Refer to the first paragraph of this proof.], so there exists a point $$x_1 \in X \setminus V_1$$. In fact, the set $$X \setminus V_1$$ is an infinite set, becuase if this set were finite, then the open covering $$\left\{ \ U_n \ \colon \ n \in \mathbb{N} \ \right\}$$ of $$X$$ would have a finite subcollection also covering $$X$$.

Suppose that $$n \in \mathbb{N}$$ and $$n > 1$$, and suppose that the $$n-1$$ distinct points $$x_1, \ldots, x_{n-1} \in X$$ have been chosen.

Using the same reasoning as in the paragraph preceding the last one, the set $$V_n = U_1 \cup \cdots \cup U_n$$ is a proper subset of $$X$$ [Refer to the first paragraph of this proof again.] and as the set $$X \setminus V_n$$ is an infinite set; so there exists a point $$x_n \in X \setminus V_n$$ such that $$x_n \neq x_j$$ for any $$j = 1, \ldots, n-1$$.

In this way we obtain the infinite subset $$S$$ of $$X$$ given by $$S \colon= \left\{ \ x_1, x_2, x_3, \ldots \ \right\}. \tag{3}$$

Furthermore we note that, for any $$n \in \mathbb{N}$$, as $$x_n \not\in V_n$$, and as $$V_n \supset V_j$$ for each $$j = 1, \ldots, n$$, by virtue of (1) above, so we can also conclude, for each $$n \in \mathbb{N}$$, the following: $$x_n \not\in V_j \mbox{ for any } j = 1 \ldots n. \tag{4}$$

Since $$X$$ is limit point compact, the infinite set $$S$$ of $$X$$ as defined in (3) above in the paragraph prior to the preceding paragraph has a limit point $$p$$ in $$X$$.

Now using (2*) above, as $$p \in X = \bigcup_{n = 1}^ \infty V_n,$$ so there exists a natural number $$r$$ such that $$p \in V_r$$; let $$r$$ be the smallest such natural number.

Now as $$p$$ is a limit point in $$X$$ of set $$S$$ and as $$V_r$$ is an open set of $$X$$ containing $$p$$, so $$V_r \cap \big( S \setminus \{ p \} \big) \neq \emptyset,$$ that is, there exists an element $$x_k$$ of $$S \setminus \{ p \}$$ such that $$x_k \in V_r$$ also.

Thus we have $$x_k \in V_r$$. But by (4) above as $$x_k \not\in V_k$$, so $$V_r \not\subset V_k,$$ and hence in view of (1) above we can conclude that $$r \not\leq k,$$ which implies that $$r > k ,$$ that is [Note that $$x_k \in S$$. Refer to (3) above.], $$k \in \{ 1, \ldots, r-1 \}.$$

Therefore the open set $$V_r$$ containing the limit point $$p$$ of set $$S$$ can intersect $$S$$ in only finitely nany points. But since $$X$$ is a $$T_1$$ space, $$V_r$$ must intersect $$S$$ in infinitely many points, by Theorem 17.9 in Munkres. Thus we have a contradiction.

Thus if a $$T_1$$ topological space $$X$$ is limit point compact, then $$X$$ is also countably compact.

Is this proof correct and clear enough in its presentation? Or, are there any problems in it of accuracy, clarity, or detail?

For the sake of contradiction, suppose that $X$ is limit-point compact but not countably compact. Then, there exists a countable open cover $(U_n)_{n\in\mathbb N}$ that admits no finite subcover. Choose $x_1\notin V_1\equiv U_1$. There exists some $n_1\in\mathbb N$ such that $x_1\in U_{n_1}$. Now choose $$x_2\notin V_2\equiv U_1\cup\ldots\cup U_{n_1}.$$ Choose $n_2\in\mathbb N$ such that $x_2\in U_{n_2}$. Then pick $$x_3\notin V_3\equiv U_1\cup\ldots\cup U_{n_1}\cup\ldots\cup U_{n_2}.$$ Pick $n_3\in\mathbb N$ such that $x_3\in U_{n_3}$. And so forth (note that $n_1<n_2<n_3<\ldots$). This way, one can construct an infinite set $D\equiv(x_1,x_2,\ldots)$ and an increasing sequence of open sets $(V_n)_{n\in\mathbb N}$ such that \begin{align*} x_m\notin V_n\quad\text{for any $m,n\in\mathbb N$ such that $m\geq n$}.\tag{$\clubsuit$} \end{align*} Note that $(V_n)_{n\in\mathbb N}$ is an open cover of $X$ as well. Also, the set $D$ infinite, since the points $x_1,x_2,\ldots$ are all distinct.

Now, if $X$ is limit-point compact, then the infinite set $D$ has a limit point $x\in X$. Let $m_0\in\mathbb N$ be such that $x\in V_{m_0}$ (remember that $(V_n)_{n\in\mathbb N}$ is an open cover). By the limit point property, there exists some $m_1\in\mathbb N$ such that $x_{m_1}\neq x$ and $x_{m_1}\in V_{m_0}$. One then must have $m_1<m_0$; see ($\clubsuit$). Since $X$ is $T_1$, the set $\{x_{m_1}\}^c$ is open and $$x\in V_{m_0}\cap\{x_{m_1}\}^c,$$ so there must exist some $m_2\in\mathbb N$ such that $x_{m_2}\in V_{m_0}\cap\{x_{m_1}\}^c$ (in particular, $x_{m_2}\neq x_{m_1}$, so $m_2\neq m_1$) and $x_{m_2}\neq x$. It follows again by ($\clubsuit$) that $m_2<m_0$. But then $$x\in V_{m_0}\cap \{x_{m_1}\}^c\cap\{x_{m_2}\}^c,$$ and so forth. Because of the limit-point property, one should be able to continue this pattern indefinitely. The problem is that one eventually runs out of distinct indices less than $m_0$, since the index set $\{1,\ldots,m_0-1\}$ is finite. Contradiction.

Your proof of the other direction seems correct to me.

$\Rightarrow$ Suppose $A\subseteq X$ has no limit points. Wlog $A$ is countable so that $A=\left \{ a_{1}, a_{2},\cdots \right \}$.

Take $U_{n}=X-\left \{ a_{n},a_{n+1},\cdots \right \}$. Each $U_{n}$ is open in $X$ since, since $\left \{ a_{n},a_{n+1},\cdots , \right \}$ is a subset of $A$, and is therefore closed in $X$ because $A$ has no limit points. It is easy to see that $X=\bigcup _{n\geq 1}U_{n}$ so that $\left \{ U_{n} \right \}_{n\geq 1}$ is a countable open cover of $X$ which has no subcover.

$\Leftarrow$ Suppose that $X$ is not countably compact and let $\left \{ U_{n} \right \}_{n\geq 1}$ be a countable open cover of $X$ which has no subcover. Then we may choose, for each $n\in \mathbb N$ an $x_{n}\in X-\bigcup^{n} _{j=1}U_{j}$ such that $x_{i}\neq x_{k}$ if $i\neq k$. Set $A=\left \{ x_{n} \right \}_{n\in \mathbb N}$. Now choose $x\in X$. Then $x\in U_{L}$ for some $L\in \mathbb N$ since the $U_{n}$ form a cover of $X$. But by construction, $x_{i}\notin U_{L}$ as soon as $i\geq L$. Therefore, $U_{L}$ s a neighborhood of $x$ that intersects $A$ in only finitely many points and so $x$ is not a limit point of $A$. As $x$ was arbitrary, $A$ has no limit points in $X$.

Note: we used the fact that if $x$ is a limit point of a subset $A$ of a $T_{1}$ space $X$ then $A\cap (N_{x}-\left \{ x \right \})$ is infinite, for every neighborhood $N_{x}$ of $x$. For, if not, then there is an $N_{x}$ neighborhood of $x$ such that $A\cap (N_{x}-\left \{ x \right \})$ is finite and hence closed in $X$. But then, $U=N_{x}-(A\cap (N_{x}-\left \{ x \right \}))$ is open in $X$, contains $x$ and its intersection with $A$ is at most $\left \{ x \right \}$, which contradicts the fact that $x$ is a limit point of $A$.