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

Question

Definition (Countable Compactness):

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 (Limit-Point Compactness):

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?

Answer 1

$\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 $A$.

Answer 2

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.

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Your proof of the other direction seems correct to me.