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I happened to come across two different versions of Bolzano-Weierstrass theorem.

Theorem $1$: Every bounded sequence in $\Bbb R^n$ has a convergent subsequence.

Theorem $2$: Every bounded infinite subset of $\Bbb R^n$ has a limit point.

The first version was found in Elementary Analysis: The Theory of Calculus by Kenneth A. Ross, while the second version was found in a lecture note of an analysis course. I would like to show that the two theorems are equivalent in the sense that we can proof one of them by assuming the other, and the proofs shall use mainly the two theorems.

Here is my attempt. Assuming Theorem $1$, consider a bounded infinite subset $S$ of $\Bbb R^n$. Use axiom of choice (or maybe countable choice) to select a countably infinite subset of $S$, call it $S_0$. By definition of countable, we have a bijection $f:\Bbb N\to S_0$. The sequence $f$ has a convergent subsequence $f'$ by assumption of Theorem $1$, and since all terms in $f$ are distinct, there is at most one term in $f'$, say $f'(m)$, such that $$\lim_{n\to\infty}f'(n)=f'(m).$$ So in fact, the limit of $f'$ is a limit point of $S_0$.

But I am having trouble with proving that Theorem $2$ implies Theorem $1$. I assumed Theorem $2$ and started with a bounded sequence $g:\Bbb N\to\Bbb R^n$. Then consider the range of $g$. If $g(\Bbb N)$ is finite, say $|g(\Bbb N)|=k$, then there is an element that appears infinitely many times in the sequence $g$, and the subsequence of only that element certainly converges. To see that such element exists, we consider the elements $c_1, c_2,...,c_k\in g(\Bbb N)$. If each of them appears finitely many times in the sequence, then there are only finitely many numbers $m$ in $\Bbb N$ such that $g(m)\in g(\Bbb N)$, a contradiction.

If $g(\Bbb N)$ is infinite, then from Theorem $2$, it has a limit point, say $L$. I suppose that $L$ is the limit of a subsequence of $g$, but I don't know how to show this. My question is the following:

Prove that an accumulation point of a sequence $x$ is the limit of a subsequence of $x$.

Any help would be appreciated.

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  • $\begingroup$ Im not sure but I think that you can get the theorem 2 without the use of any choice. I think that using the nested interval theorem to prove theorem 2 is enough and dont need the use of the axiom of choice. Anyway Im not completely sure. $\endgroup$ – Masacroso Sep 22 '16 at 11:48
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    $\begingroup$ @Masacroso I suppose that by nested interval theorem, you mean Cantor intersection theorem. I would like to avoid using Cantor intersection theorem or Heine-Borel theorem, and simply use each of Theorem 1 and 2 to prove the other. $\endgroup$ – edm Sep 22 '16 at 11:56
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    $\begingroup$ @Masacroso Yes, you don't need the axiom of choice. Instead you split your bounded area into $n$ (open but not necessarily disjoint) unit intervals/squares and pick any interval with infinite many points in it. Take interval again and repeat the procedure (making the intervals smaller). For every interval choose an arbitrary point in it. This sequence converges by Bolzano and you can show this limit is an accumulation point (since every open set covers an interval around that point which contains infinitely many points). $\endgroup$ – ctst Sep 22 '16 at 11:58
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I found a solution just now. We first prove a lemma concerning about open balls of $L$.

Lemma $1$: For all $\varepsilon>0, B_\varepsilon(L)\cap g(\Bbb N)\setminus \{L\}$ is an infinite set.

The intersection operator and set difference operator are associative in this case, so I don't need to add parentheses.

Proof: For a particular $\varepsilon>0$, $B_\varepsilon(L)\cap g(\Bbb N)\setminus \{L\}$ is guaranteed to be nonempty. Choose any element in this set, say $y_1$. Notice that $\varepsilon_1:=d(L,y_1)>0$, and we choose any element $B_{\varepsilon_1}(L)\cap g(\Bbb N)\setminus \{L\}$, say $y_2$. We have that $\varepsilon_2:=d(L,y_2)>0$, and get yet another nonempty set $B_{\varepsilon_2}(L)\cap g(\Bbb N)\setminus \{L\}$. We can repeat this process indefinitely due to axiom of choice. Now we have countably infinite number of elements $y_1,y_2,...$ picked from $B_\varepsilon(L)\cap g(\Bbb N)\setminus \{L\}$ , and they are all distinct due to how they are picked. So in fact, $B_\varepsilon(L)\cap g(\Bbb N)\setminus \{L\}$ is an infinite set.

We now proceed to prove that $L$ is the limit of a subsequence of $g$. We find the subsequence by "truncating" a sequence. We define $\varepsilon_n=\frac{1}{n}$ for each $n\in\Bbb N$, and find any element in $B_{\varepsilon_1}(L)\cap g(\Bbb N)\setminus \{L\}$, call it $g(m_1)$. Then, we consider the subsequence of terms that appear after $g(m_1)$, i.e. we consider $\{g(x):x>m_1\}$. There are still elements in this set that are also in $B_\varepsilon(L)\setminus \{L\}$ because we removed at most finitely many elements from an infinite set (here, Lemma $1$ is used). We pick any element in $\{g(x):x>m_1\}\cap B_{\varepsilon_2}(L)\setminus \{L\}$, call it $g(m_2)$. Repeat the construction by picking $g(m_k)$ from $\{g(x):x>m_{k-1}\}\cap B_{\varepsilon_k}(L)\setminus \{L\}$ for each $k\in\Bbb N$. Due to how the sequence $g(m_1),g(m_2),...$ is constructed, it is a subsequence of $g$, and converges to $L$.

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You do the same trick as before, when your sequence only had finite range. Since $L$ is an accumulation point you get for every $n \in \mathbb{N}$ infinitely many $x \in g(\mathbb{N})$ s.t. $\|x-L\|<1/n$ holds. Since there are infinitely many of those, you get for every $N$ a $i(n)>N$ s.t. $\|g(i(n))-L\|<1/n$ holds. Setting $N=i(n-1)$ in the thinking before and using this subsequence you are done.

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  • $\begingroup$ It seems to be a bad idea writing $\|x_n-L\|<1/n$, with the same $n$ on both sides. For example, the sequence $1/\sqrt n$ has $0$ as a limit point, but $1/\sqrt n>1/n$ for all but one terms. $\endgroup$ – edm Sep 22 '16 at 13:00
  • $\begingroup$ @ctst in such a construction, you are not ensuring that the subsequence's index is increasing, I mean you can form a subsequence like "$n_5, n_3, n_6, ...$" with this construction. $\endgroup$ – onurcanbektas Oct 27 '17 at 7:37
  • $\begingroup$ @onurcanbektas You are ensuring this here: $i(n) >N=i(n-1)$ $\endgroup$ – ctst Oct 27 '17 at 10:15
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Consider the set of pairs $(\frac1n,x_n)$ with some product metric. This is obviously bounded and has exactly one point for each element of the sequence.

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