Can we discard 'bounded' from the hypothesis?

Question

From baby Rudin:

'If $\{E_n\}$ is a sequence of closed nonempty bounded sets in a complete metric space $X$, if $E_n\supset E_{n+1}$, and if $\lim\limits_{n\to \infty} \text{diam}\, E_n=0$, then $\cap_1^{\infty} E_n$ consists of exactly one point.'

Can we discard 'bounded' from the hypothesis, because $\lim\limits_{n\to \infty} \text{diam}\, E_n=0$ implies bounded, doesn't it?

Answer 1

Specifying bounded means that the diameter is always finite. Having infinite diameter meaning "unbounded" is no particular obstacle, but it means that there are small issues with writing $\lim_{n\to \infty} \operatorname{diam} E_n$. What do limits mean when taken from the extended reals? What if the sequence takes the value $\infty$ infinitely many times? (I know that nestedness implies that this is impossible, but for more general extended real sequences.)

I haven't read Baby Rudin, but I'm guessing that, like most real analysis texts, the analysis is done over the real numbers, not the extended real numbers. It's not that any of these issues are insurmountable (on the contrary, it's quite straightforward to iron them out), but that requires work - arguably unnecessary work - to say the same thing.

It's much simpler just to specify that the sets are bounded. If any of them are unbounded, but the sequence eventually becomes bounded, people can just omit the unbounded sets from the sequence.

Answer 2

I agree with you that $$\lim\limits_{n\to \infty} \text{diam}\, E_n=0$$

implies boundedness and also if one of $E_n$ is bounded then the nestedness of the sequence implies that all $E_m$ for $m>n$ are bounded as well.

Answer 3

$\lim\limits_{n\to \infty} \text{diam}\, E_n=0$ implies that there is $N \in \mathbb N$ such that $\text{diam}\, E_n <1$ for all $n>N$. Hence, if $n >N$, then $E_n$ is bounded.