Which of these topological properties imply which?

Question

I am going through the chapter on compactness and completeness from Sternberg's Advanced Calculus and trying to build an intuition for what many of this topological properties mean, and which imply which. The book defines these concepts in the setting of metric spaces, but most of what I found online is in the about topology, and from what I see (correct me if I am wrong) it doesn't change the general picture much.

I've made this diagram to see whats the relationship of the different concepts and have examples of each. I don't know if it is correct. For example, Is it true that a bounded complete metric (sub)space is compact (and therefore totally bounded)? Then why bother defining total boundedness? If you don't think anything else I wrote is a valid question stick to answering that, although pointing out any misconceptions I might have is appreciated.

For intuition about compactness I've found this posts really helpful. It helps me to think that there are (at least) two different kinds of infinity: one in the sense of largeness (of which boundedness is the opposite), and another in the sense of denseness (of which discreteness is the opposite).

Answer 1

I have not read Sternberg's Advanced Calculus but I think that it might be wrong about "boundedness + completeness implies compactness".

For example take any infinite set, say the unit interval $I=[0,1]$. Now consider the discrete metric on it; that is, consider the metric space $(I,d)$ where $d(x,y)= 1$ if and only if $x=y$ and $d(x,y)= 0$ otherwise.

By definition, $I$ is compact if every open cover of it admits a finite subcover. Cover $I$ with the collection of open balls $\mathcal{C} = \{U_x\}_{x \in I}$, where $U_x = B_{1/2}(x) = \{x\}$. It is clear, by the infinitude of $I$, that $\mathcal{C}$ does not admit a finite subcover and then $I$ isn't compact. However $I$ is bounded, as the maximum value of $d$ is $1$, and $(I,d)$ is complete, as every cauchy sequence is eventually constant(to see this take $\epsilon = 1/3$ in the definition of cauchy sequence).

The last paragraph proves that boundedness and completeness does not imply compactness. Furthermore, the same example (with the same cover $\mathcal{C}$) shows that boundedness and completeness together does not imply the Lindelöf property. Lindelöf property can be thought of as a weakening of compactness requiring every open cover to have a countable subcover. See this article.

In contrast with the previous counterexample, it's true that total boundedness and completeness together imply compactness. Thus, the point of defining total boundedness will be to have a "metric space charactrization" of compactness.

Answer 2

Just to clear up an apparent misconseption. If we have the same definition of "closed subset", then whether a topological (or metric) space is closed or not is not a well defined property without specifying which superspace the property is measured against. For example, every space is a closed subset of itself. So just saying that $\mathbb{Q}$ is closed is meaningless.

Also, "in an infinite dimensional normed vector space" is a seemingly trivial property. Whatever set (of cardinality less than or equal to $\mathbb{R}$) you give me, I can find an infinite dimensional normed vector space to embed it in.