Clarifications regarding the reals

[Bachelier]

We know $$\mathbb{R} \ or \ \mathbb{R}^1$$ is complete, hence every Cauchy seq. converges. Therefore every subsequence of them converges as well. (to the same pt)

My clarification is regarding the term Sequentially Compact (SC). In the definition, they only state a topological space X is SC if every sequence in X has a convergent subsequence.

I know there is the term compact in the name but it is never mentioned in the definition. Only later do we make the connection between compactness and SC via the theorem: X is SC iff it is compact. Which begs the question that the definition of sequentially compact has nothing to do with compactness in the first place.

So if it doesn't and the relationship between the two properties is only proven later after introducing both terms separately, then based on these definitions, $$\mathbb{R}$$ is sequentially compact. And therefore it is compact per theorem, which we know is not true because it is unbounded.

 

[jambaugh]

The sequence 1,2,3,4,... in R had no convergent subsequence.

 

[HallsofIvy]

No, R is NOT "sequentially compact" for two reasons.

1) It is not true that "every sequence of real numbers has a convergent subsequence". For example, the sequence {1, 2, 3, 4, ...} of all positive integers has no convergent subsequence. What is true is that every bounded sequence of real numbers has a convergent subsequence.

2) You have stated the definition of "Sequentially Compact" slightly incorrectly: a set is "sequentially compact" if and only if every sequence of points in that set converges to a point in the set. That will be the same as saying "every sequence has a convergent subsequence" if the set is closed.

That is, every set of real numbers will be "sequentially compact" if and only if it is both closed and bounded, which, of course, also implies "compact".

 

[Bachelier]

Great. Thank you guys. I knew I was misreading the definitions.