# (ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed.

Let $X$ be a metric space. Let $\{p_n\}$ be a sequence in $X$. Let $E$ be a set of all subsequential limits of $\{p_n\}$.

How do i prove that $E$ is closed in ZF?

Is there a well-ordering of convergent subsequences? I can't think of one..

• If I understand the question correctly, what you mean is: How can one prove this without using the axiom of choice? – Michael Hardy Aug 16 '12 at 16:42
• @tomasz: If the sequence is finite then it is closed already; otherwise the sequence itself is a countable dense subset. So $E$ is Dedekind-infinite, and a separable metric space on its own accord. – Asaf Karagila Aug 16 '12 at 16:43
• Do you know it is true (ie, is it an exercise, or has someone told you it is true?) – Thomas Andrews Aug 16 '12 at 16:45
• @Thomas It's not an exercise. The theorem is in PMA Rudin, and the argument used AC. I don't know whether this is provable in ZF – Katlus Aug 16 '12 at 16:48
• For any $x\in E$ you can define the sub-sequence $p_{i_1},...,p_{i_n},...$ inductively with $i_n$ the least number satisfying $i_n>i_{n-1}$ and $d(p_{i_n},x)<\frac{1}{n}$. Then given a sequence $x_1,..,x_n,...\in E$, with $x_n\to x$, you can do a "diagonalized" version of this construction, I believe to construct a sub-sequence which converges to $x$ - define $i_n$ to be the least number satisfying $i_n>i_{n-1}$ and $|x_n-p_{i_n}|<\frac{1}{n}$ – Thomas Andrews Aug 16 '12 at 16:48

## 3 Answers

Let us denote $\{ p_n : n \in \mathbb{N} \}$ by $B$.

Clearly $x \in \overline{ B }$ iff $x = p_n$ for some $n$, or $( \forall m ) ( \exists n ) ( 0 < d (x,p_n ) \leq \frac{1}{m} )$. In the latter case we can construct, via induction and without any choice, a subsequence converging to $x$. (In fact, the latter condition is easily seen to be equivalent to $x$ being a subsequential limit of $(p_n)_{n \in \mathbb{N}}$.)

Suppose $x \in \overline{ B }$ is not a subsequential limit. This means that $x = p_n$ for some $n$. But also that there is an $m \in \mathbb{N}$ such that $d ( x , p_n ) > \frac{1}{m}$ for all $n$. Therefore $x$ is an isolated point of $B$, and thence it is also an isolated point of $\overline{ B }$.

It thus suffices to show that $\overline{ B } \setminus \{ x : x\text{ is an isolated point of }B \}$ is closed, and this follows from the following:

Claim: Suppose $F \subseteq X$ is closed and $A \subseteq F$ is a set of isolated points of $F$. Then $F \setminus A$ is closed.

Proof: It suffices to show that $X \setminus ( F \setminus A)$ is open. If $x \in X \setminus ( F \setminus A )$ there are two cases: Either $x \notin F$, in which case $x \in X \setminus F$, and this is a neighbourhood of $x$ disjoint from $F \setminus A$. Otherwise $x \in A$, but as $x$ is isolated there is a neighbourhood $U$ of $x$ such that $F \cap U = \{ x \}$, and therefore $( F \setminus A ) \cap U = \emptyset$.) $\dashv$

Edit: There is a minor issue in what I have done above. I seem to have assumed that the sequence $( p_n )_{n \in \mathbb{N}}$ is one-to-one (silly me). This means that some of what I said above is not quite true in general. The relevant facts we still have are:

• If $x \in \overline{B}$ then either $x \in B$ or there is a subsequence of $( p_n )_{n \in \mathbb{N}}$ converging to $a$.
• If there is a subsequence of $( p_n )_{n \in \mathbb{N}}$ converging to $x$, then either $( \forall m ) ( \exists n ) ( 0 < d (x,p_n ) \leq \frac{1}{m} )$, or $( \forall N ) ( \exists n \geq N ) ( p_n = x )$.

The set we wish to show is closed is therefore $\overline{B} \setminus A$ where $$A = \{ x \in B : x\text{ is an isolated point of }A\text{ and }( \exists N ) ( \forall n \geq N ) ( p_n \neq x ) \}.$$ As $A$ is a set of isolated points, by the above $\overline{B} \setminus A$ is closed.

• I understand that $\overline{ B } \setminus \{ x : x\text{ is an isolated point of }B \}$ is closed. However, i don't understand why this means "every limit point of 'a set of all sequantial limits' is in the set". Help – Katlus Aug 16 '12 at 18:25
• @Katlus: Do you see how every non-isolated point of $B$ is a limit of a subsequence of $( p_n )_{n \in \mathbb{N}}$? – user642796 Aug 16 '12 at 18:31
• Yes, i was also talking about isolated points. Let $E$ be a set of all sequential limits. Then $\overline{ B } \setminus \{ x : x\text{ is an isolated point of }B \} \subset E$, but converse doesn't seem to hold.. – Katlus Aug 16 '12 at 18:55
• @Katlus: I think I may have handled it in my edit, above. If $x \in E$, then either $x$ is the limit of a non-constant sequence, or there is a constant subsequence with values $x$. If the former case ever holds, the $x \in \overline{B}$ and cannot be an isolated point. If the former case never holds, then $x$ must appear infinitely often in the sequence, and I (no longer) remove it. – user642796 Aug 16 '12 at 19:28
• Now i got it fully. Nice construction! – Katlus Aug 16 '12 at 19:39

Check if this is ok: $\forall$ $y$ in the closure of $E$, let $N_{y}$ be any neighbourhood of $y$. Then, by definition of an element of the closure, for any neighbourhood of $y$, $N_{y} \bigcap E \neq \emptyset$.

By the same token, for any z in E, $N_{z}$ (any neighbourhood of z) is such that $N_{z} \bigcap \left(p_{n}\right) \neq \emptyset$.

Therefore, there is a $z \in N_{y} \bigcap E$ so that $N_{z} \bigcap \left(p_{n}\right) \neq \emptyset$.

Let $N = N_{z} \bigcap N_{y}$. Then N is a neighbourhood of z and $N \subset N_{y}$ (intersections of neighbourhoods are neighbourhoods and z is in both)

Therefore $N_{y} \bigcap \left(p_{n}\right) \neq \emptyset$. And $y$ is in $E$, so that $E$ is closed, since $N_{y}$ is arbitrary

Since in this construction there are only intersections of sets, I think it is independent from AC.

How about my proof?

proof: Let $(p_n)$ be a sequence in metric space $X.$ If $q\in X$ is a subsequential limit of sequence $(p_n),$ then there exists a subsequence $(p_{\sigma_q(m)})_{m\in\mathbb{N}}$ of $(p_n),$ such that $\lim\limits_{m\to\infty} p_{\sigma_{q}(m)}=q,$ where $\sigma_{q} : \mathbb{N}\to\mathbb{N}$ is increasing strictly. Put $$E=\{q\in X\mid \exists \text{ subsequence }(p_{\sigma_q(m)})_{m\in\mathbb{N}} \text{ of (p_n)_{n\in\mathbb{N}} } : \lim\limits_{m\to\infty} p_{\sigma_q(m)}=q \}.$$ To show $E$ is closed, let $x\in E',$ and we shall prove that $x\in E.$

For $n=1,$ since $\hat{\mathbb{B}}(x,\frac{1}{2\cdot 1})\cap E\neq\emptyset,$ pick $q_1\in\hat{\mathbb{B}}(x,\frac{1}{2\cdot 1})\cap E,$ then $q_1\in E$ and $d(q_1,x)<\frac{1}{2}.$ Hence there exists subsequence $(p_{\sigma_1(m)})_{m\in\mathbb{N}}$ such that $\lim_{m\to\infty} p_{\sigma_1(m)}=q_1.$ Hence there exists $M_1\in\mathbb{N}$ such that for all $m\geq M_1, d(p_{\sigma_1(m)}, q_1)<\frac{1}{2},$ thus $d(x,p_{\sigma_1(M_1)})\leq d(x,q_1)+d(q_1, p_{\sigma_1(M_1)})<1.$ For $n=2,$ since $\hat{\mathbb{B}}(x,\frac{1}{2\cdot 2})\cap E\neq\emptyset,$ pick $q_2\in\hat{\mathbb{B}}(x,\frac{1}{2\cdot 2})\cap E,$ then $q_2\in E$ and $d(q_2,x)<\frac{1}{2\cdot 2}.$ Hence there exists subsequence $(p_{\sigma_2(m)})_{m\in\mathbb{N}}$ such that $\lim_{m\to\infty} p_{\sigma_2(m)}=q_2.$ Hence there exists $M_2\in\mathbb{N}$ with $M_2>\sigma_1(M_1)$ such that for all $m\geq M_2,$ $d(p_{\sigma_2(m)}, q_2)<\frac{1}{2\cdot 2},$ thus $$d(x,p_{\sigma_2(M_2)})\leq d(x,q_2)+d(q_2, p_{\sigma_2(M_2)})<\frac{1}{2}$$ and $\sigma_2(M_2)\geq M_2>\sigma_1(M_1)$ (note that $\sigma_2$ is increasing strictly). Now assume we have constructed $p_{\sigma_1(M_1)}, \dots, p_{\sigma_k(M_k)},$ such that \begin{gather*} d(x, p_{\sigma_i(M_i)})<\frac{1}{i}, \qquad \forall i=1,\dots,k \end{gather*} with $\sigma_1(M_1)<\cdots<\sigma_{k}(M_k).$ As for $n=k+1,$ Since $x\in E',$ $\hat{\mathbb{B}}(x,\frac{1}{2(k+1)})\cap E\neq\emptyset.$ Pick $q_{k+1}\in \hat{\mathbb{B}}(x,\frac{1}{2(k+1)})\cap E,$ then $q_{k+1}\in E$ with $d(q_{k+1}, x)<\frac{1}{2(k+1)}.$ Hence there exists subsequence $(p_{\sigma_{k+1}(m)})_{m\in\mathbb{N}}$ such that $\lim\limits_{m\to\infty} p_{\sigma_{k+1}(m)}=q_{k+1}.$ Hence, there exists $M_{k+1}\in\mathbb{N}$ with $M_{k+1}>\sigma_{k}(M_k)$ such that for all $m\geq M_{k+1},$ $d(p_{\sigma_{k+1}(m)}, q_{k+1})<\frac{1}{2(k+1)}.$ Thus \begin{gather*} d(x,p_{\sigma_{k+1}(M_{k+1})})\leq d(x,q_{k+1})+d(q_{k+1}, p_{\sigma_{k+1}(M_{k+1})})<\frac{1}{k+1}, \end{gather*}
with $\sigma_{k+1}(M_{k+1})\geq M_{k+1}>\sigma_{k}(M_k).$ Hence, let $\eta: i\mapsto\sigma_i(M_i)\in \{p_n\mid n\in\mathbb{N}\}, \forall i\in\mathbb{N},$ then the mapping $\eta : \mathbb{N}\to\mathbb{N}$ is increasing strictly, and the principle of induction follows that we have constructed a subsequence $(p_{\eta(n)})_{n\in\mathbb{N}},$ such that \begin{gather*} d(x,p_{\eta(n)})<\frac{1}{n},\qquad \forall n\in\mathbb{N} \end{gather*} holds. Since $1/n\to 0,$ as $n\to\infty,$ we see that $\lim\limits_{n\to\infty} p_{\eta(n)}=x,$ hence $x\in E.$ Since $x$ is chosen arbitrary, we conclude that $E$ is closed.

Note that we write $\mathbb{B}(x,r):=\{y\in X\mid d(x,y)<r\}$ and $\hat{\mathbb{B}}(x,r)=\mathbb{B}(x,r)\backslash\{x\}.$