Can a convergent sequence take on non-finite values in one direction? Baby Rudin 3.2 c)

Question

In Baby Rudin 3.2, there is a proof for the following:

Theorem. If $\{p_n\}$ converges, then $\{p_n\}$ is bounded.

Proof. Suppose $p_n \rightarrow p$. There is an integer $N$ s.t. $n>N$ implies >$d(p_n, p) < 1$. Put $$r=max\{1, d(p_1,p), d(p_2, p),...d(p_N,p)\}$$ Then $d(p_n, p)<r$ for $n=1,2,3,...$.

I was wondering why there couldn't be a sequence that stretches infinitely far in the negative direction that eventually converges to a point.

I think it might be because I'm still a bit shaky on what can belong to a metric space. It says $\{p_n\}$ is a sequence in metric space X, but can the metric only take on finite values? What's to stop a sequence from taking on non-finite values?

Any help will greatly be appreciated!

Answer 1

I was wondering why there couldn't be a sequence that stretches infinitely far in the negative direction that eventually converges to a point.

The idea of $\{p_n\}$ converging to some limit $p$ is that for large $n$, all $p_n$ will be sufficiently close to $p$.

In other words, given $D > 0$, only finitely many $p_n$ will fail to satisfy $d(p_n, p) < D$ and this is why the function won't "stretch infinitely far in some direction".

By definition, $d(x, y)$ is always a finite real number.

Answer 2

"stretches infinitely far in one direction" isn't a well-defined notion.

If for every integer $z \in \mathbb{Z}$ there's some point $p_n$ such that $p_n \le z$, then you can construct a subsequence of $\{p_n\}_{n \in \mathbb{N}}$ which diverges to $-\infty$ and hence the sequence isn't convergent.