Theorem 3.2 in Rudin's book states that any convergent sequence $\{p_n\}$ in a metric space $X$ is bounded. I understand everything he did, but it looks "incomplete" to me.
Paragraphing his proof:
Suppose $p_n \to p$, so there exists $N$ so that $n \geq N$ implies $d(p_n, p) < 1$. Set $r = \max\{1, d(p_1, p), \ldots, d(p_N, p)\}$. So $d(p_n, p) \leq r$ for all $n$.
The proof ends here. Rudin's definition of boundedness is "a sequence is bounded if its range is bounded." Range here, I believe, means $\{p_n\}$, i.e., regarding a sequence as a set, which makes sense to me. How does showing this proof that the range of the sequence is bounded?
Best Answer
To be precise, given a sequence $(p_n)_{n=1}^{\infty}$, the range is defined to be $$R = \{p_n: n \ge 1\}\text{.}$$ We then say that the range $R$ (noting this is a set) is bounded if, according to definition 2.18(i), given a metric space $X$, there is a real number $M$ and a point $q \in X$ such that $d(p, q) < M$ for all $p \in R$.
In the proof, with a metric space $X$, we took an arbitrary $p_n \in R$ and a $p \in X$ and found a value $r$ so that $d(p_n, p) \leq r$. One can thus choose any real number greater than $r$, say $s$, and state that $d(p_n, p) < s$. Hence $R$ is bounded.