Proof check that the set $A$ of subsequential limits of a series ${p_n}$ in a metric space is closed.

Question

Let $q$ be an accumulation point of $A$. There exists a point $x$ such that $d(x.q)< \frac{\delta}{2}- d(p_i,p_{m_j}) \tag{1}$ where $p_{m_{j}}$ is an arbitrary element of the subsequence {${p_{m_i}}$} which converges to $x$ (such a sequence exists since $x$ is in $A$) which means that for any $\epsilon>0$ there exists an integer $M$ such that whenever $m_i>M$, $d(p_{m_i},x)<\epsilon$. Setting $\epsilon=\frac{\delta}{2}+d(p_i,p_{m_j})$ we get $d(p_{m_i},x)< \frac{\delta}{2}+d(p_i,p_{m_j}) \tag{2}$ Adding $(1)$ and $(2)$ we get $$d(x,q)+d(p_{m_i},x)<\delta$$ using the triangle inequality, $$d(q,p_{m_i})<\delta$$ whenever $m_i>M$. So the subsequence ${p_{m_i}}$ converges to $q$ and it is in $A$. Is this correct?

Answer 1

No, it is not correct. For instance, you don't say what is that $\delta$ that you introduce in the second sentence of the proof. And it makes no sense to mention $d(m_i,x)$, since $m_i$ is an integer, not an element of your space.

Answer 2

Remark: I see that Apostol's accumulation point is different from limit point due to his word distinct. Then, by a definition, a set is closed iff it contains all of its accumulation points (in the sense of Apostol).

I understand that (perhaps) you want to prove that

Theorem   The set $\ A\ $ of all accumulation points (in the sense of Apostol) of any subset $\ S\ $ of arbitrary metric space $\ X\ $ is closed in $\ X$.

In particular, $\ S\ $ may be the set of all terms of a sequence in $\ X,\ $ for $\ S\ :=\ \{p_n: n\in\mathbb N\},\ $ where $\ p: \mathbb N\rightarrow X\ $ is an arbitrary sequence.

You may try to write a proof in a natural way by starting routinely (automatically) by expanding definitions:

Let $\ B\ $ be the set of all accumulation points of $\ A\ $. We want to show that $\ B\subseteq A.\ $ Thus, let point $\ b\in B.\ $ Every open ball $\ U(b,\ r)\ (\mbox{where}\ r>0)\ $ contains a point $\ a\in A\setminus\{b\}.\ $ Then, in turn, there exists $\ x\in U(a,\,\ d(a,\ b))\cap S\setminus\{a\}.\ $ Etc.

Be careful, you have to show that $\ x\ne b.$

Good luck.

Answer 3

I feel that there is a need for a clean presentation of the unhappy situation created by combining and discussing together the two notions:

• closed set
• accumulation point

(The hidden nuisance problem is due to logical element "distinct". This forces considering two cases where a single homogenous one would do).

At the given elementary level, it is necessary to spell out the relevant definitions. Everything happens within a metric space $\ (X\ d).\ $ Our OP have already kindly provided the definition of the accumulation point of arbitrary set $\ S\subseteq X.\ $ Now, by definition:

DEFINITION $\ S\ $ is closed $\ \Leftarrow:\Rightarrow\ A\subseteq S,\ $ where $\ A\ $ is the set of all accumulation points of $\ S.$

Now, under the above notation, when $\ A\ $ is the set of all accumulation points of $\ S,\ $ the OP problem can be sated as:

THEOREM   $\ A\ $ is closed.

PROOF Let $\ B\ $ be the set of all accumulation points of $\ A.\ $ According to the above definition of a closed set, we have to show that

$$ B\subseteq A $$

If you read my previous answer all the way to "Etc." you will see that every point $\ b\in B\ $ is an accumulation point of $\ S\ $ hence $\ b\in A;\ $ in other words, $\ B\subseteq A.\ $ End Of PROOF

I felt it pedagogical to write these two notes instead of one, it should help -- that's how I feel.