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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.

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  • $\begingroup$ 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. $\endgroup$ Feb 14 '18 at 20:21
  • $\begingroup$ If in $(x_n)$ the points get further from $x$, then $x$ isn't a limit point of $A$. $\endgroup$ Feb 14 '18 at 20:22
  • $\begingroup$ 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. $\endgroup$
    – H_1317
    Feb 14 '18 at 20:47
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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 $B(x, \frac{1}{n})$).

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  • $\begingroup$ Great argument on the infinite points being available and we've only used finite many at each step ... on the infinite point intersection result, not sure I saw this formally stated in my book -- would it be because we can pick some radius r and find some point inside that radius that belongs to A at distance d from the center of our ball. then we can shrink our radius to less than d and find another point that intersects this new subset of our original ball? and so on we can continue to shrink our radius using every real number at our disposable -- yielding infinite points in the intersection $\endgroup$
    – H_1317
    Feb 25 '18 at 0:32
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    $\begingroup$ @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. $\endgroup$ Feb 25 '18 at 16:06

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