Is this a valid example of a metric space that is bounded but not totally bounded?

Question

Looking up examples of spaces that are bounded but not totally bounded, I came across some complex examples (in Banach spaces, etc). I have attempted to construct a simpler one.

Is the following an example of a bounded but not totally bounded space:

We take the metric space $\Bbb{R^2}$. Let us consider the sequence $\{a_i\}$ along the x-axis defined thus: $a_k=(\frac{1}{2^k},0)$. Clearly this sequence is convergent to $(0,0)$. Now let each of the points in $\{a_i\}$ be the limit of a sequence along the y-axis. For example, the sequence converging to $(\frac{1}{2^k},0)$ can be $(\frac{1}{2^k},\frac{1}{2^j})$ for all $j\in\Bbb{N}$. The points are clearly bounded.

For any $\epsilon\in\Bbb{R}$, we are to construct a finite set $J$ such that $\bigcup_{i=1}^n B(j_i,\epsilon)$ covers all the points given. As there are infinite sequences converging to points of the form $(\frac{1}{2^k},0)$, and $J$ is finite, $J$ can contain points from only a finite number of them. As for the selected points, $\epsilon$ can be made small enough so that no points from the sequences from which no points are selected are included in the balls $B(j_i,\epsilon)$. Hence, the space is bounded, but not totally bounded.

Is this example correct? Thanks in advance!

Answer 1

Let your space be $X$. Your argument fails because for any $a>0$ we can choose $J$ to contain all of the points of $X$ in the square $[a,1]\times[a,1]$: there are only finitely many points in that square. The remaining points of $X$ just don’t cover much space, and it’s not too hard to see how to cover them with small balls centred at just finitely many of them.

Make sure that the origin is also in $j$, and $B\big(\langle 0,0\rangle,\sqrt2a\big)$ covers every point of $X$ in $[0,a)\times[0,a)$, leaving only the parts of $X$ in $[a,1]\times[0,a)$ and $(0,a)\times[a,1]$ to be covered. Pick any $x\in(0,a)$; it’s not hard to see that there’s a finite subset $F$ of $\big(\{x\}\times[a,1]\big)\cap X$ such that balls of radius $\sqrt2a$ centred at points of $F$ cover $\big([0,a)\times[a,1]\big)\cap X$. Finally, the finitely many balls of radius $\sqrt2a$ centred at points of $X$ in $[a,1]\times\{0\}$ cover $\big([a,1]\times[0,a)\big)\cap X$. Altogether, then we have a finite set of points such that the balls of radius $\sqrt2a$ centred at those points covers $X$, and we can certainly make $\sqrt2a$ as small as we like.

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Here’s another fairly explicit way to see that $X$ is totally bounded. You know that $[0,1]\times[0,1]$, being compact, is totally bounded. Given $\epsilon>0$, let $F_\epsilon$ be a finite subset of $[0,1]\times[0,1]$ such that $$\mathscr{B}=\left\{B\left(x,\frac{\epsilon}2\right):x\in F_\epsilon\right\}$$ covers $[0,1]\times[0,1]$. Clearly $X\subseteq[0,1]\times[0,1]$, so $\mathscr{B}$ covers $X$. The problem, of course, is that $F_\epsilon$ need not be a subset of $X$, but we can get around this.

For each $x\in F_\epsilon$ define a point $y_x\in X$ as follows: if $B\left(x,\frac{\epsilon}2\right)\cap X\ne\varnothing$, let $y_x$ be any point of $B(x,\epsilon)\cap X$, and otherwise let $y_x$ be the origin. Let $D_\epsilon=\{y_x:x\in F_\epsilon\}$; clearly $D_\epsilon$ is a finite subset of $X$. Suppose that $p\in X$. Then $p\in B\left(x,\frac{\epsilon}2\right)$ for some $x\in F_\epsilon$, so $B\left(x,\frac{\epsilon}2\right)\cap X\ne\varnothing$, and $y_x\in B\left(x,\frac{\epsilon}2\right)\cap X$. This implies that $d(p,y_x)\le d(p,x)+d(x,y_x)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$, i.e., that $p\in B(y_x,\epsilon)$. Since $p$ was an arbitrary point of $X$, $\{B(y,\epsilon):y\in D\}$ covers $X$, and $X$ is therefore totally bounded.

Answer 2

Your method are doomed to fail. Here is why:

1. Totally bounded is equivalent to the condition that the space have finite cover each with radius less than $\epsilon$ for any $\epsilon>0$.

2. Metric subspace of a totally bounded metric space is also totally bounded. Trivially prove from above.

3. By Heine-Borel theorem: every closed and bounded subset in $\mathbb{R}^{n}$ is compact.

4. Every compact metric space is totally bounded.

5. The closure of any bounded subset is bounded.

6. Combine all those above, every bounded subset in $\mathbb{R}^{n}$ is a subset of its closure. The closure is both closed and bounded, and hence compact, and hence totally bounded. Thus the subset is also totally bounded.

Therefore, any example constructing from a subset of $\mathbb{R}^{n}$ is doomed to fail.

That's probably why all examples you found come from infinite dimensional space. You don't need anything fancy really. Even the sequence space $l^{2}(\mathbb{R})$ is good enough.

EDIT: to prove 1:

Left to right implication: for any $\epsilon$ cover the space with a finite bunch of ball with radius $\frac{\epsilon}{3}$.

Right to left implication: for any $\epsilon$ cover the space with a finite cover each with radius at most $\frac{\epsilon}{2}$. Then pick any point for each set in the cover and produce a ball of radius $\epsilon$, which will be enough to cover the set.

Answer 3

The problem is here:

"As there are infinite sequences converging to points of the form $\left(\frac{1}{2^k},0\right)$, and $J$ is finite, $J$ can contain points from only a finite number of them."

Here $J$ is finite, but the balls you mention will contain finally all the terms of all the sequences converging to that limit point.