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?
Best Answer
$\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 $A$.