Hey there! I totally get why this part feels confusing—Rudin has a habit of glossing over small but critical details that feel "obvious" once you connect the dots, but are tricky to unpack at first glance. Let’s walk through exactly why that constructed infinite subset has no limit point in $R^k$, which is the key gap in understanding the (c) ⇒ (a) proof.

Quick Recap of the Proof Strategy #

To prove (c) implies (a), we use proof by contradiction: we assume E does NOT satisfy (a) (so either E is unbounded, or E is not closed) and show this violates condition (c) (i.e., we can find an infinite subset of E that has no limit point in E). Your confusion centers on the unbounded case, so let’s focus there.

The Unbounded Case: Building the Problematic Subset #

If E is unbounded in $R^k$, by definition, for every positive integer $n$, there exists a point $x_n \in E$ such that $|x_n| > n$ (here $|\cdot|$ is the Euclidean norm in $R^k$). We claim the infinite subset $S = {x_n \mid n \in \mathbb{N}}$ has no limit point in $R^k$—let’s prove this rigorously.

Why $S$ Has No Limit Point in $R^k$ #

Suppose, for contradiction, that there exists some $y \in R^k$ that is a limit point of $S$. By the definition of a limit point:

For every $\varepsilon > 0$, there are infinitely many $x_n \in S$ such that $|x_n - y| < \varepsilon$.

Now use the triangle inequality:
$$|x_n| \leq |x_n - y| + |y|$$
Rearranging gives:
$$|x_n - y| \geq |x_n| - |y|$$

Since $|x_n| > n$, substitute that in:
$$|x_n - y| > n - |y|$$

Pick $\varepsilon = 1$ (any positive $\varepsilon$ works, but 1 is straightforward). Choose an integer $N > |y| + 1$. For all $n > N$, we have:
$$|x_n - y| > n - |y| > (|y| + 1) - |y| = 1 = \varepsilon$$

This means only finitely many points of $S$ (those with $n \leq N$) can satisfy $|x_n - y| < \varepsilon$. But this directly contradicts the definition of a limit point, which requires infinitely many such points. Therefore, our assumption that $y$ exists is false—$S$ has no limit point in $R^k$.

Why This Breaks Condition (c) #

Since $S$ is an infinite subset of E, and it has no limit point in $R^k$, it certainly has no limit point in E. This violates condition (c), which states every infinite subset of E must have a limit point in E. So our initial assumption that E is unbounded is wrong—E must be bounded.

For completeness, if E were not closed, we’d take a limit point $z$ of E that’s not in E, construct an infinite subset of E converging to $z$, and show this subset has no limit point in E (since $z \notin E$), which also violates (c). But your main confusion was the "no limit point in $R^k$" step, so that’s the core breakdown.

内容的提问来源于stack exchange，提问作者Vinicius L. Deloi