# Understanding Rudin's proof of: Every bounded sequence in $R^k$ contains a convergent sequence

I am trying to understand the proof for:

Theorem 3.6b: Every bounded sequence in $$R^k$$ contains a convergent subsequence

which is as follows in Baby Rudin:

This follows from (a), since Theorem 2.41 implies that every bounded subset of $$R^k$$ lies in a compact subset of $$R^k$$.

(a) refers to:

If $$\{p_n\}$$ is a sequence in a compact metric space $$X$$, then some subsequence of $$\{p_n\}$$ converges to a point of $$X$$

and Theorem 2.41 is:

2.41 $$\ \$$ Theorem $$\ \$$ If a set $$E$$ in $${\bf R}^k$$ has one of the following three properties, then it has the other two:

$$\quad(a)\ \$$ $$E$$ is closed and bounded.
$$\quad(b)\ \$$ $$E$$ is compact.
$$\quad(c)\ \$$ Every infinite subset of $$E$$ has a limit point in $$E$$.

My interpretation of Rudin's proof of Theorem 3.6b is:

Proof: Any bounded sequence of $$R^k$$ is clearly contained in a k-cell and since each k-cell is compact (Rudin's Theorem 2.40), every bounded sequence of $$R^k$$ lies in a compact subset of $$R^k$$. Now, let $$\{p_n\}$$ be an arbitrary sequence in $$R^k$$. Then, by Theorem 3.6a, some subsequence of $$\{p_n\}$$ converges to a point of $$R^k$$ and we are done.

My question: Is my interpretation of the proof correct? Is it the interpretation that Rudin was pointing to?

• I also agree with you, too! Feb 4, 2022 at 13:05

2.42$$\quad$$Theorem (Weierstrass)$$\quad$$ Every bounded infinite subset of $$R^k$$ has a limit point in $$R^k$$.
Proof$$\quad$$ Being bounded, the set $$E$$ in question is a subset of a $$k$$-cell $$I \subset R^k$$. By Theorem 2.40, $$I$$ is compact, and so $$E$$ has a limit point in $$I$$, by Theorem 2.37.
Here, Theorem 2.37 states that "If $$E$$ is an infinite subset of a compact set $$K$$, then $$E$$ has a limit point in $$K$$."