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?
Best Answer
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.