Comparison of sequential compactness and limit point compactness. I'm studying analysis, but I want to know more concepts about metric spaces, so I try to read some explanations in topology books. My textbook is Munkres' Topology (second edition). I cite some definitions used in this book for convenience.


*

*Compact: Every open covering $\mathcal{A}$ of $X$ contains a finite subcollection that also covers $X$.

*Limit point compact: Every infinite subset of $X$ has a limit point.


And after giving these two definitions, the author proved that for any space, Compactness ⇒ Limit point compactness, but not conversely.
And on p.179, the author said that

We now show these two versions of compactness coincide for metrizable spaces; for this purpose, we introduce yet another version of compactness called sequential compactness.



*Sequentially compact: Every sequence of points of $X$ has a convergent subsequence.


Then the author stated a theorem that these all three definitions are equivalent in the metric space case. He prove this fact in the order: (1)⇒(2)⇒(3)⇒(1), where (1)⇒(2) is mentioned just now, (2)⇒(3) is easy; however, the (3)⇒(1) part, as the author said, is the hardest part of the proof, that used Lebesgue number lemma.
Now my question one is, I have seen a book that gave a quite easy proof for (2)⇒(1); that is to say, we can prove Compactness ⇒ Limit point compactness (in the metric space case).
So we don't need to introduce the term Sequentially compact in order to prove this equivalence right? But the author said,

We now show these two versions of compactness coincide for metrizable spaces; for this purpose, we introduce yet another version of compactness called sequential compactness.

This confused me. Or is it the fact that the Sequentially compactness itself deserves to have a place anyway? That is my question one.

My second question is: are limit point compactness and sequentially compactness equivalent in general spaces? How about in a metric space? Does the proof of this go simple when in metric space case? I see some books did not even have a phrase of the Limit point compactness, why?

My third question is, what does the Bolzano-Weierstrass Theorem truly say? Here I list some candidates:


*

*Every bounded infinite set in $\mathbb{R}^n$ has a limit (accumulation) point in $\mathbb{R}^n$. (Apostol)

*Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. (Many analysis books write so.)

*The closed and bounded subset of $\mathbb{R}^n$ is a limit point compact space. (Suggested by me.)

*The closed and bounded subset of $\mathbb{R}^n$ is a sequentially compact space. (Suggested by me.)


Yet, in the history that real analysis just developed, the original Bolzano-Weierstrass Theorem may be stated like the first two type.
However, now that the Topology has studied a lot, in pedagogically aspect, which of the description of the theorem is better to think and to use nowadays?
 A: Def'n 1. is the general def'n  of compactness in topology, whether or not the topology can be generated by a metric.
In a  metric space, a set $X$ is compact iff every sequence in $X$ has a limit point that BELONGS to $X$ iff every infinite subset of $X$ has a limit point that belongs to $X.$ 
The n-dimensional generalization of the (1-dimensional) Bolzano-Weierstrass theorem is that a subset of $R^n$ is compact iff it is closed and bounded.  (But this does not hold for all metric spaces.)
It is useful to know various equivalents to compactness, and to know that there are equivalents specific to metric spaces, just as it is  useful to know various equivalents  of "continuous function", some of which are specific to metric spaces. 
Another term is "pre-compact" which I have seen only in the context of Hausdorff spaces : $X$ is pre-compact iff $\overline X$ is compact.  
A: Sequential compactness implies limit point compactness in general. Let S be an infinite set. Since S is infinite you can have a sequence such that $x_n \neq x_m  $ for all n,m integer. Sequential compactness implies this guy has a convergent subsequence and limit x of this subsequence is a limit point of set S since every neighborhood of x contains a tail of the subsequence(so there are infinitely many of them) it contains one element other than x . (In this prove I assume axiom of choice)
At first glance one may think they can prove limit point compactness implies sequential compactness with following idea. let $x_n$ be a sequence then $S= {x_n}$ is either finite than it has converging subsequence trivially so nothing to prove or is infinite and the set has limit point so I can find a subsequence converge to it but this is not true.
First try to construct sequence to see the problem your self
As an e
The example I found in  the book "counterexamples in topology" is real numbers with right order topology. This topology whose basis are $(a,\infty)$
consider singleton set ${a}$ if $x<a$ every element of basis containing x will contain {a} thus every set containing an element has limit point.Sequence $x_n = - n$ given an open set $(b,\infty)$ there exist N s.t -nN. So sequence will escape any open set eventually ,thus, sequence cannot have a convergent subsequence.
