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In the article about metric space on wikipedia:

If $X$ is a complete subset of the metric space $M$, then $X$ is closed in $M$. Indeed, a space is complete if it is closed in any containing metric space.

Do the second sentences say that:

In a metric space, a subset is complete if and only if the subset is closed in the metric space.

If not, what is the relation between completeness and closedness for a subset in a metric space?

Thanks and regards!

Martin Sleziak
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Tim
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  • I have not seen the phrase "complete subset" before & I hope not to see it again. The article is vague. It should say that if $X,$ with the metric $d$ inherited metric $d$ of $ M,$ is a complete metric space, then $X$ is closed in $M.$...We must distinguish between a metric space and a metrizable space. A metric space is a triplet $(M,d,T)$ where $T$ is the topology on $M$ generated by the metric $d$.... A metrizable space is a pair $(X,T)$ where $T$ is a topology on $X$ such that there exists a metric $d$ on $X$ for which $(X,d,T)$ is a metric space..... (continued in next comment). – DanielWainfleet Mar 03 '18 at 08:33
  • If $(M,d,T_M)$ is a metric space and $X\subset M$ then there may be a complete metric $e$ on $X$ that generates the same topology on $X$ that $d$ does even if $X$ is not closed in $M.$ That is, let $T_{X,d}$ be the topology on $X$ generated by $d.$ There may be a $complete$ metric $e$ on $X$ (not on $ M$) such that $T_{X,e}=T_{X,d}$ even if $d$ (restricted to $X$) is not a complete metric on $X.$ ... A topological space whose topology can be generated by some complete metric is called a completely-metrizable space. – DanielWainfleet Mar 03 '18 at 08:45

2 Answers2

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No; if that were the case, then $\mathbb{Q}$ would be complete, since it is closed in itself.

What the second sentence says is:

Let $X$ be a metric space. Then $X$ is complete if and only if for every metric space $M$, if $X$ is contained in $M$ then $X$ is closed in $M$.

In particular, if you have a fixed complete $X$ and a fixed metric space $M$ that contains $X$, then you can conclude that $X$ is closed in $M$ (the first sentence).

Arturo Magidin
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Notice: A subset is complete iff it is closed in any metric space in which it is a subspace.

A closed subspace of a complete metric space is complete. On the other hand, a closed subspace of a metric space need not be complete. Take the space $(0,1)$ which is closed in itself. But it is not complete. So closedness is not enough for completeness.

However a compact metric space is always complete. Compactness is the right property.

For this and more, try out G. F. Simmons, "Introduction to Topology and Modern Analysis".

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    I'm not sure what you mean by saying "compactness is the right property"; not every complete metric space is compact (e.g., the reals). Presumably, you mean a property that is sure to guarantee completeness? – Arturo Magidin Aug 17 '10 at 18:41
  • Yes, that is what I mean. –  Aug 17 '10 at 19:04