Distance between two points in a totally bounded set in a Banach space is finite?

Question

I am doing the proof that every totally bounded set in a Banach space is bounded. The fact that a set is totally bounded means that for every $\epsilon >0$ there are points $x_1,...,x_n$ such that $B_\epsilon (x_j)$, $j = 1,...,n $ is a covering of the set.

My question is if it is possible to have a single isolated point that is infinitely far away from the other points? Say I have for example a totally bounded set $U$ that is contained in $ B_R (0)$, $0<R< \infty$ (so $\forall x \in U: ||x||<R$) except for one point $a \in U, ||a|| = \infty $. Then $||a-x|| = \infty$ for every $x \in U, x \neq a$ and any $\epsilon$ ball centered at $a$ would cover $a$. Is such a situation possible?

Answer 1

In any metric space (and that includes Banach spaces), the distance of two points is always finite by definition.

I guess the confusion may come from something like this: if $X$ is the space of all functions from $\mathbb{R}$ to $\mathbb{R}$ with the supremum norm, and $f$ is defined by $f(x)=x$ for all real $x$, then $\Vert f\Vert=\infty$. However, this is exactly the reason why $X$ isn't a normed space, since there are elements of infinite norm and, by definition, it doesn't induce a metric. (If you restrict it to bounded functions, it will be a Banach space). I hope this helps to clear it up a little bit.

Answer 2

Its impossible for $\|x\| = \infty$ for any $x$, it has to be a non-negative real number by definition.

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To show that a totally bounded set $U$ is bounded, you can just take some epsilon, say $\varepsilon = 1$.

We know that $U\subseteq \bigcup_{i=1}^n B(x_i, 1)$ for some $x_i$. To show that $U$ is bounded, we need to show it contained in some open ball. For this we can take $x_1$ as its center. Take $r = 1+\max_{j = 1, ..., n} \|x_1-x_j\|$ which is one more than the maximal distance between $x_1$ and other points from $x_1, ..., x_n$.

If $y\in B(x_i, 1)$ then $\|x_1-y\| \leq \|x_i-y\|+\|x_i-x_1\| < 1+\|x_i-x_1\| \leq r$, so $y\in B(x_1, r)$.

Hence $U\subseteq \bigcup_{i=1}^n B(x_i, 1)\subseteq B(x_1, r)$.