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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?

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

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Yes, your interpretation is perfectly fine, well done!


I'll also add that it is interesting to compare and contrast this result with Theorem 2.42 and its proof (given on page 40):

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 $K$."

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