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?
Proof check that the set $A$ of subsequential limits of a series ${p_n}$ in a metric space is closed.
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amathproblemthatneedsolve
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Your [MATH]LaTeX[/MATH] isn't working for me. A common problem I'm ever to familiar with. see if the link helps. https://oeis.org/wiki/List_of_LaTeX_mathematical_symbols Otherwise just click the button above and insert you math symbols the button says "add LaTeX Powered by MathJax"
Let [MATH]q[/MATH] be an accumulation point of [MATH]A[/MATH]. There exists a point [MATH]x[/MATH] such that [MATH]d(x.q)< \frac{\delta}{2}- d(p_i,p_{m_j}) \tag{1}[/MATH]} where [MATH]{p_{m_j}}[/MATH] is an arbitrary element of the subsequence [MATH]{p_{m_i}}[/MATH] which converges to [MATH]x[/MATH] (such a sequence exists since[MATH] $x$[/MATH] is in[MATH] A[/MATH]) which means that for any [MATH]\epsilon>0[/MATH] there exists an integer [MATH]M[/MATH] such that whenever [MATH]m_i>M[/MATH], [MATH]d(p_{m_i},x)<\epsilon[/MATH]. Setting [MATH]\epsilon=\frac{\delta}{2}+d(p_i,p_{m_j})[/MATH] we get [MATH]d(p_{m_i},x)< \frac{\delta}{2}+d(p_i,p_{m_j}) \tag{2}[/MATH] Adding [MATH](1)[/MATH] and [MATH](2)[/MATH] we get [MATH]d(x,q)+d(p_{m_i},x)<\delta[/MATH] using the triangle inequality, [MATH]d(q,p_{m_i})<\delta[/MATH] whenever [MATH]m_i>M[/MATH]. So the subsequence [MATH]{p_{m_i}}[/MATH] converges to q and it is in A. Is this correct?
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pka
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Use [ tex][/tex] without the first space.
Will please simply state what is to be proved,
pka
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What is rude reply. The title is truly incoherent showing lack of any understanding of standard definitions, Because you are working in a metric space there is more structure with which to work, But definitions vary. The term accumulation point can and does vary in different texts. It can have the same meaning as limit point or cluster point. Moreover, you use the term "proof check" which has no meaning as well as saying limits of a series. Well a series is a summation which has at most one limit point. I think you must have intended to use the term sequence.
So given the total muddle of the title, I simply ask that you state clearly what you were to prove.
I suspect that you are asked to show that the set of all accumulation points of a set is a closed set.