# Is every Closed set a complete space?

a closed set contains all its cluster (accumulation) points: points for which any open neighborhood around them no matter how small contains points from the set.
a complete set contains all limit points of Cauchy sequences. which are very similar to cluster points.
My question is: in a metric space and a closed interval (usual topology) is it possible to have points clustering without a cluster point?

Closed sets and cluster points are concerns of topology and topology is independent of the chosen metric. Saying the space is metric does not factor into topological considerations. How familiar are you with the ideas of modern topology and manifolds?

I know they are topological considerations, but a metric space is a topological space and it has special properties, for example a metric space is "normal" which is not true for a general topological space. Im just wondering if in a metric space with a topology generated by open balls if closed sets and completeness properties are related . just like an open interval (EDIT: on the Real line) is not a complete metric space yet a closed interval would contain all its limit points..

I'm not so familiar, i just have been opening some books (M.Nakahara and P. Szekeres ) which are beyond my level, a lot of this I've seen yesterday for the first time..

Thanks!

dextercioby
Homework Helper
The reverse is true independent of the chosen topology, while the doubt about the direct statement/question relies in the possibility that some Cauchy sequences wrt the metric topology are not convergent wrt the same topology. I'm sure a mathematician could come up with a Cauchy sequence which is not convergent wrt some metric topology, so that a closed set is not always a complete space.

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If a subset of a metric space is complete, then the subset is always closed. The converse is true in complete spaces: a closed subset of a complete space is always complete.

An example of a closed set that is not complete is found in the space $X=\mathbb{R}\setminus \mathbb{Q}$, with the usual metric. Then X is a closed set of itself but is not complete.
Curiously, there exists a metric on X such that X does become complete (and such that the topology doesn't change). So we say that X is complete metrizable (i.e. there exists a metric such that the metric space is complete, this does NOT mean that the space is complete for every metric).

One can say that a closed set of a complete metrizable space is complete metrizable. But there is not much more we can say.

A space that is not complete metrizable is $\mathbb{Q}$.

Wait a minute. By metric space, you mean Riemannian manifold, don't you? If not, then I have no idea.

dextercioby