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So I am defining the set convergence of a sequence $\{a_n\} \subset \mathbb{R}$ as follows:

$$ L = \{p \in \mathbb{R} \quad | \quad \exists \{b_n\} \subset \{a_n\} \quad b_n \longrightarrow p\} $$ and I note this as $a_n \longrightarrow L$

My question is:

Is it true that for any subset $A$ of $\mathbb{R}$ there is some $\{a_n\} \subset \mathbb{R}$ such that $a_n \longrightarrow A$?

Axel
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    It's true if and only if $A$ is a closed set. – bof Jan 21 '21 at 07:41
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    Any closed set $A$ of real numbers is the closure of some countable set $D$. Now consider an infinite sequence in which each element of $D$ occurs infinitely often. – bof Jan 21 '21 at 07:45

1 Answers1

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It is not true. If $L$ is the limit of a sequence, then $L$ is closed. This is easiest to see from a topological point of view. Your condition for set convergence can be reformulated this way:

$a_n$ converges to the set $L$ if for every point $p\in L$, every open neighborhood of $p$ contains elements of $a_n$.

Now let $x$ be a boundary point of $L$, that is, every open neighborhood $U_x$ of $x$ intersects $L$. Let $y\in U_x\cap L$. Now since $U_x$ is open, there is an open neighborhood $U_y$ of $y$ contained in $U_x$. And since $y\in L$, this neighborhood contains elements of $a_n$. But then so does $U_x$, since it's a superset of $U_y$. Since $U_x$ was an arbitrary neighborhood, this holds for all neighborhoods, and thus $x\in L$. So $L$ must contain all of its boundary points and is thus closed. Unless a set is closed, it can't be the limit of a sequence.

Uncountable set limits are possible, though. Enumerations of the rationals have the reals as their limit, since the rationals are dense in the reals.

Also, $L$ being closed is a sufficient condition for being the limit set of a sequence. Take a dense, countable subset of $L$ and enumerate it with the sequence $a_n$. Using this, define the sequence $b_n$ which goes like this:

$$a_1,~a_1,a_2,~a_1,a_2,a_3,~a_1,a_2,a_3,a_4,~a_1,a_2,a_3,a_4,a_5,~\dots$$

You get the pattern. Now any point in $L$ is a limit point of $b_n$, since $a_n$ enumerates a dense subset, meaning that all points in $L$ are limits of sequences whose elements are elements of $a_n$, and any such sequence can be constructed as a subsequence of $b_n$. No points outside of $L$ are limit points, since $L$ is closed and thus already contains all limit points of any sequence contained in $L$ (which $b_n$ is by construction).

Vercassivelaunos
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    Why don't you just say that a set $L$ is the set of all subsequential limits of a sequence if and only if it's a closed set? – bof Jan 21 '21 at 07:40
  • @bof: Because I couldn't think of a nice proof for the reverse direction in the moment. I added a proof for the reverse direction, but I'm still not too happy with it, since it takes a not too small amount of manual constructions instead of using structural arguments. Do you have an idea for a nicer proof? – Vercassivelaunos Jan 21 '21 at 11:45
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    A nonempty countable set can be enumerated enumerated in a sequence in which each element is repeated infinitely often; there is no need to specify the enumeration in detail. – bof Jan 21 '21 at 12:27