# Complete metric space in which the large balls are not compact.

Find an example of a complete metric space $$X$$ in which all sufficiently small closed balls are compact but large ones are not.

First of all, what is ''a small ball'' and ''a large ball'' in a metric space? Does it have to do with the radius of the ball?

Theorem: Every closed ball of $$X$$ is compact $$\Leftrightarrow$$ $$X$$ has the nearest-point property.

Since $$X$$ has not the nearest-point property, it is not compact. But $$X$$ is complete so the space is not total bounded.

I think this is the fact that we want to take advantage of, to prove that large balls are not compact.

Not sure how to proceed or if this thinking is correct.

The nearest-point property is not in many topology books, so i give the definition:

$$(X,d)$$ is a metric space. The following are equivalent:

i) Every infinite bounded subset of $$X$$ has an accumulation point in X(BW criterion)

ii) Every bounded sequence of $$X$$ has a convergence subsequence in $$X$$.

iii) X is complete and every bounded subset of $$X$$ is totally bounded.

Let $$X$$ be an infinite discrete space with the usual discrete metric, $$d(x,y)=1-\delta_{xy}$$. Since $$X$$ is infinite and discrete $$\{x\}_{x\in X}$$ is an open cover with no finite subcover. Cauchy sequences in $$X$$ are eventually constant, and thus converge, and any sufficiently small closed ball contains a single point and is thus compact. Note that $$X$$ is the closed ball of radius 1 around any point. Hence $$X$$ satisfies your conditions.

Edit: Some comments on my thought process, in response to the question in the comments.

My first thought was that we know that a metric space is compact if and only if it is complete and totally bounded. Since our whole metric space is complete here, we need to find a way to make small closed balls totally bounded and large balls not totally bounded. The way I think about total boundedness, this is similar to saying we want small neighborhoods of a point to be "finite dimensional" and large neighborhoods to be "infinite dimensional." In a very rough intuitive way.

From there, the discrete topology is fairly natural, since small balls are literally finite and large balls contain the whole space.

As for my thoughts on your thoughts, well, they're correct, but I'm not sure where you would go from where you got to.

• Thank you very much for your answer sir! Can you please explain to me how did you think about the problem? Can you comment on my thinking? – argiriskar Dec 28 '18 at 20:00
• @argiriskar edited. – jgon Dec 28 '18 at 20:17

Here is a simple way to construct a metric space such that $$B_r(x)$$ is compact for $$r<\varepsilon$$ and not compact for $$r\geq\varepsilon$$, for a fixed $$\varepsilon>0$$. Let $$(X,d)$$ be a noncompact unbounded metric space in which closed balls are compact (such as $$\Bbb R$$ with the usual metric $$d(x,y)=|x-y|$$) and consider $$d'(x,y)=\min\{d(x,y),\varepsilon\}$$. It can be shown that $$d'$$ is a metric and it induces the same topology as $$d$$ (even though the latter is not needed for your question). Now closed $$d'$$ balls of radius $$<\varepsilon$$ are compact since they are the same as closed $$d$$ balls, but balls of radius $$\geq\varepsilon$$ are not compact, since they are the whole space.

• Of course you can also take the discrete distance scaled by $\varepsilon$, but my example highlights how boundedness and being a ball are really properties of the metric, not of the topology it induces, which can be useful to keep in mind – Alessandro Codenotti Dec 28 '18 at 20:52