Limit point compactness implies sequentially compact

I have a question on one of the steps of the following proof, from Munkres Topology, page 180:

Let X be a metrizable space. Then if X is limit point compact, then X is sequentially compact.

Proof:

Consider the set A = {$x_n$ | n $\in$ Z+ }, where ($x_n)$ is some sequence of points in x.

In the case where A is infinite:

If A is infinite, then A has a limit point x. We define a subsequence of ($x_n$) converging to x as follows:

First choose $n_1$ so that

$$x_n{_1} \in B(x,1)$$

Then suppose that the positive integer $n_{i-1}$ is given. Because the ball B(x,1/i) intersects A in infinitely many points, we can choose an index $n_i$ > $n_{i-1}$ such that

$$x_n{_i} \in B(x,1/i)$$

Then the subsequence $x_n{_1}$, $x_n{_2}$, ... converges to x.

My question is:

How do we know we can find a subsequence via the above process that maintains the correct order that the original sequence had? For example, maybe we choose our first element, $x_1$, inside the disk of radius 1, but then as we go on choosing elements we are unable to find an element within a smaller disk of radius less than 1 because those elements were listed before $x_1$ in our original sequence, A.

As an example, I was thinking what if our sequence was constructed as all points in X with the order of lowest to highest distance from x. Then the earliest points in the sequence, ($x_n$), would be the closest points to x and as we move on in the sequence the points get farther way from x.

Then in the proof above, we would pick some $x_n{_1}$ to start within the disk of radius one, but as we shrink our disk radius, we look to move along our sequence ($x_n$) to find a point that falls within our shrinking radius. However as we move along our sequence we will eventually find points that fall outside of our shrinking radius.

Thanks in advance; I'm sure I've gone wrong somewhere.

• At each stage, there are infinitely many admissible points, but only finitely many can occur before the point of the sequence you are currently at. Feb 14 '18 at 20:21
• If in $(x_n)$ the points get further from $x$, then $x$ isn't a limit point of $A$. Feb 14 '18 at 20:22
• i believe it still would be a limit point of A, i've just taken every open disk around x, and ordered the points at the boundry of the disk lowest to highest distance wise in my sequence. Feb 14 '18 at 20:47

In a metric space, $x$ being a limit point of $A= \{x_n : n \in \mathbb{N}\}$ means that for every $r>0$, $B(x,r) \cap A$ is infinite. So if we have chosen $x_{n_1} \in B(x,1)$, we have finitely many points of $A$, namely $\{x_1, \ldots, x_{n_i}\}$ we can ignore, and we then consider $r=\frac{1}{2}$, and $B(x,r)$ intersects $A$ in an infinite set, so certainly in $A \setminus \{x_1, \ldots, x_{n_1}\}$, and we pick some $x_n$ from that to get $n_2$, and we can continue this inductively, as at each $i$ we can ignore all lower indexed $x_n$ (so for $n < n_i$) and pick $x_{n_{i+1}} \in B(x, \frac{1}{i+1})$ from among the other points so that automatically $n_{i+1} > n_i$. So we get an increasing subsequence, as required.
The only part of being metric we really use is that $X$ is $T_1$ so limit points have this infinite intersection property, and that the limit point has a local countable base (that we can choose decreasingly, as the balls $B(x, \frac{1}{n})$).
• @H_1317 by contradiction is easier. Suppose that $x$ is a limit point of $A$ and that for some $r>0$ we have that $F:=B(x, r) \cap A$ is finite. Then we pick $0<r’ <r$ smaller than $\min\{d(x,p): p \in F, p\neq x\}$ and note that $B(x, r’)\cap A$ is either empty or $\{x\}$, contradicting that $x$ is a limit point of $A$. So all intersections are infinite. Feb 25 '18 at 16:06