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I have learned that $X$ is a closed set if and only if every limit point is in $X$. Let a closed but incomplete set be $X=[0,3]\bigcap\mathbb{Q}$. But in this case, the limit point of the rational sequence $\{x_n=(1+\frac{1}{n})^n, n\in\mathbb{N}\}$ is in fact $e \notin \mathbb{Q}$. Why do we still say it is closed?

On the other hand, more frequently the completeness is used to describe a metric space, rather than a set. But the term "closed" describes both set and space. Can we say a set that is complete?

Analysis Newbie
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    It is closed in $\mathbb{Q}$, where $\mathbb{Q}$ has the subspace topology, so this is pretty much by definition it is closed in $\mathbb{Q}$. It certainly is not closed in $\mathbb{R}$, because take for example: ${q \in \mathbb{Q} \cap [0,1] | q^2 < 1/2}$, the supremium of this set is not in $X$ but is in in $\mathbb{R}$. – Maithreya Sitaraman Sep 24 '17 at 19:01
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    Do we say that it's closed? Closed where? – Noah Schweber Sep 24 '17 at 19:01
  • @Analysis Newbie What is your definiton of limit point? – S.S.Danyal Sep 24 '17 at 19:10
  • The most fundamental definition of closed set is that its complement is an open set. This set is neither open nor closed in $\mathbb{R}$

    see https://math.stackexchange.com/questions/1476978/can-a-set-be-neither-open-nor-closed

     – WW1 Sep 24 '17 at 19:18
  • It’d be more conventional to notate your sequence as $$x_n = \left{ \left( 1+\frac1n \right) ^n \right}_{n\in\Bbb N}$$ – gen-ℤ ready to perish Sep 24 '17 at 20:04
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    $[0,3]\cap \mathbb Q$ is closed in topologies where it is closed and not in togologies where it is not. In the metric space $\mathbb R$ it is not closed for precisely the reason you give. But in the metric space $\mathbb Q$ it closed because every limit point of the set is a member of the set. The point $e$ is not a limit point of the set, because $e$ simply does does not exist in $\mathbb Q$ so it can not be talked about. It simple does not exist. – fleablood Sep 25 '17 at 02:48

2 Answers2

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First, "set" and "space" means the same thing unless you put some other structures.

For example, we say the x-axis $$\{(x,0):x\in \mathbb{R}\}$$ is a subspace of $\mathbb{R}^2$, and we don't use subset here is because this subspace is the linear subspace (closed under addition and scalar multiplication).

In a topological space or metric space, set and space means the same thing, and you can just think of subset and subspace both as subset.

Now we move to closed and complete.

  1. Closed. It does not make sense to say a set is closed. In general when we say $[0,1]$ is closed, we are actually saying $[0,1]$ is closed in $\mathbb{R}$. Closed only makes sense if we have a subset $A\subset X$, and we can say $A$ is closed in $X$, and using your words, this means all the limit point of $A$ in $X$ are actually in $A$. Using your example, note $A:= [0,1]\cap \mathbb{Q}$ is closed in $\mathbb{Q}$, you can check all its limit point in $\mathbb{Q}$ are actually in $A$, and as you observed $A$ is not closed in $\mathbb{R}$.

  2. Complete. It makes sense to say a set is complete just by itself: a set is complete means that it is closed in all of its superset. So a set $A$ is complete if it is closed in each $X$ where $X\supset A$. Intuitively, you can think about completeness as a stronger notion of "closed". In your example, $A:= [0,1]\cap \mathbb{Q}$ is not complete, because $A$ is not closed in $\mathbb{R}\supset A$.

Xiao
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  • I disagree with your definition of "complete," as it only applies when the parent topological spaces in question are Hausdorff. For example, $\mathbb{R}$ is complete, but if you define $X = \mathbb{R}\cup {a}$ to have the topology of $\mathbb{R}$, but such that every neighborhood of $0$ contains $a$, then $\mathbb{R}$ is not closed in $X$. – Michael L. Sep 25 '17 at 13:26
  • You are right, the most general case that I normally see completeness being mentioned is topological vector space. Then, let us say our space is Hausdorff. And completeness itself is not really suited for general topological space since it is not a topological property, i.e. homeomorphism does not preserve completeness. – Xiao Sep 25 '17 at 17:38
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To be a limit point of a set, a point must actually exist and be a point in the space.

So if our space is $\mathbb Q$ with the euclidean metric then....

$e = \lim\limits_{n\to \infty}(1 + \frac 1n)^n$ is not a point in $\mathbb Q$. So $e$ does not exist and can not be a limit point of any set in the space $\mathbb Q$.

$e$ does not exist.

That's all there is to it. It is that simple.

.......

Now if our space were $\mathbb R$ with the euclidean metric then that would be a different story.

Then the set $[0,3]$ would be closed; $e$ would exist; $e$ would be a limit point; and $e$ would be in the set.

But the set $[0,3]\cap \mathbb Q$ would not be closed; $e$ exists; $e$ is a limit point, and $e$ is not in the set. So the set is not closed.

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Note: In the above answer, I am stating, without any justification, that all limit points of $[0,3]\cap \mathbb Q$ in the space $\mathbb Q$ are in the set and that all the limit points of $[0,3]$ in the space $\mathbb R$ are in $[0,3]$.

Those can be proven but I thought they weren't pertainent to the answer.

fleablood
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