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Below is a theorem from these notes. I have two questions:

  1. Why can we make $d(p_{n_1},w) < 1$? I think this is a typo and what is meant is $d(p_{n_1},x)<1$. (And so $d(p_{n_1},w) \le 2$.) Is my vision correct?
  2. Why do we need to define $\delta$? Can I just replace $\delta$ with $1$ everywhere without making any other corrections? I think this will only make the proof easier to understand. Or do I miss something?

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user557
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    For your first question the answer is it is possible to choose $n_1$ such that $d(p_{n_1},w)<1$. Since $d(x,w)<1$ we can choose $n_1$ such that $d(p_{n_1},x)<1-d(x,w)$ and then triangle inequality gives $d(p_{n_1},w)\leq d(p_{n_1},x)+d(x,w)<1$. For the second question It appears to me that you are right: $\delta =1$ seems to work just fine. – Kavi Rama Murthy Dec 25 '17 at 05:18

3 Answers3

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Regarding your two questions:

  1. I think this is not a typo and it is intentionally writen $d(p_{n_1},w)<1$, since, $(p_n)$ being a subsequence converging to $x$ means that it can get arbitrarily close to $x$ and, since $x$ belongs to the unitary open ball around $w$, evidently we can find a term of $(p_n)$ inside that 1-ball around $w$ - it is open and $x\in N_1(w)$, so there exists an $r>0$ such that $N_r(x)\subseteq N_1(w)$. So, this explanation might just have been ommitted for reasons of clarity simplicity.
  2. Well, actually, any consant value would be adequate since, what matters is that denominator $2^j$ that is of exponential rank. However, $\delta$ is the most elegant choice so as to ensure some kind of monotonicity for our subsequence.
Vassilis Markos
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This is not really an answer, but might provide another perspective.

Let $E$ be the set of subsequential limits.

Suppose $z \notin E$. Consider the sets $Z_n = B(z, {1 \over n})$. Let $I = \{ n | Z_n \text{ intersects } E \}$. I claim that $I$ is finite, in which case we see that $E^c$ is open and so $E$ is closed.

Suppose $I$ is infinite. Then there is a sequence $e_n \in E$ such that $e_n \to z$. We can assume that $d(e_n,z) < {1 \over n}$ without loss of generality.

Now we need to find a subsequence of $p_n$ that converges to $z$ for a contradiction.

Each $e_n$ is a subsequential limit, so for any $N$ we can always find an index $k_n \ge N$ such that $d(p_{k_n}, e_n) < { 1\over n}$. Choose $k_{n+1} > k_n$ to create the sequence $p_{k_n}$.

Then we have $d(z,p_{k_n}) \le d(z,e_n)+(p_{k_n}, e_n) < {2 \over n}$ and so $z \in E$, a contradiction.

copper.hat
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This is not really an answer, but might provide another perspective.

The set of subsequential limits can be written as $$\displaystyle \bigcap_{n=0}^\infty \overline{\{p_{n+k}\mid k\in \mathbb N\}}$$ which is an intersection of closed sets, hence closed.

Gabriel Romon
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