Proof: By Induction

Since the real sequence $$(a_n)_{n\in\mathbb N}$$ is bounded, there are real numbers $$A,B\in\mathbb R$$ with

$A\le a_n \le B\quad\quad\forall n\in\mathbb N.$

We will prove the theorem in three steps:

Step $$1$$ - We construct intervals $$[A_k,B_k]$$ fulfilling certain properties

We will prove by induction that there exists a sequence $$(I_k)_{k\in\mathbb N}$$ of closed real intervals $$I_k:=[A_k,B_k]$$ with the following properties:

1. $$[A_k,B_k]$$ contains infinitely many sequence members of the sequence $$(a_n)_{n\in\mathbb N}$$, i.e. $$a_n\in I_k$$ for infinitely many $$n\in\mathbb N$$.
2. $$[A_k,B_k]\subseteq [A_{k-1},B_{k-1}]$$.
3. $$B_k - A_k = 2^{-k}(B-A)$$.

Base case $$k=0$$

For $$[A_0,B_0]:=[A,B]$$ the properties are obviously fulfilled.

Induction step $$k\rightarrow k + 1$$.

Let $$[A_k,B_k]$$ be an interval, for which the properties are fulfilled. Set $$M:=(A_k + B_k)/2$$. $$M$$ is the "middle" of the interval $$[A_k,B_k]$$. Since $$[A_k,B_k]$$ contains infinitely many members of the sequence $$(a_n)_{n\in\mathbb N}$$, at least one of the intervals $$[A_k,M]$$ and $$[M, B_k]$$ must also contain infinitely many members of this sequence. We set

$[A_{k+1},B_{k+1}]:=\cases{[A_k,M],\quad\text{ if }[A_k,M]\text{ contains infinitely many sequence members of }(a_n)_{n\in\mathbb N},\\ [M, B_k],\quad\text{ else.} }$

By construction, $$[A_{k+1},B_{k+1}]$$ fulfills the properties mentioned above.

Step $$2$$ - We construct a subsequence $$(a_{n_k})_{k\in\mathbb N}$$ with $$a_{n_k}\in[A_k,B_k]$$ for all $$k\in\mathbb N$$

Now, again by induction, we define the members of the subsequence $$(a_{n_k})_{k\in\mathbb N}$$ with $$a_{n_k}\in[A_k,B_k]$$ for all $$k\in\mathbb N$$ as follows:

Base case $$k=0$$

We set $$a_{n_0}:=a_0$$.

Induction step $$k\rightarrow k + 1$$.

Let the sequence members $$a_{n_0}, a_{n_1},\ldots,a_{n_k}$$ be already defined. Because the interval $$[A_{k+1},B_{k+1}]$$ contains infinitely many sequence members, there exists an $$a_{n_{k+1}}\in[A_{k+1},B_{k+1}]$$ with $$n_{k+1} > n_k$$. Thus we can include $$a_{n_{k+1}}$$ to the sequence $$a_{n_0}, a_{n_1},\ldots,a_{n_k},a_{n_{k+1}}$$.

Step $$3$$ - We prove that the subsequence $$(a_{n_k})_{k\in\mathbb N}$$ is convergent.

Let $$\epsilon > 0$$ be an arbitrarily small (but fixed) real number. Since the length of the $$k$$-th interval is $$B_k - A_k=2^{-k}(B-A)$$, there is an $$N(\epsilon)\in\mathbb N$$, for which the length the $$N(\epsilon)$$-th interval is

$B_{N(\epsilon)} - A_{N(\epsilon)}=2^{-{N(\epsilon)}}(B-A) < \epsilon.$

By construction of the intervals, for all $$k,j > N(\epsilon)$$ will have $\begin{array}{rcl} a_{n_k}\in[A_{n_k},B_{n_k}]\subseteq [A_{N(\epsilon)},B_{N(\epsilon)}],\\ a_{n_j}\in[A_{n_j},B_{n_j}]\subseteq [A_{N(\epsilon)},B_{N(\epsilon)}]. \end{array}$ Thus, for all $$k,j > N(\epsilon)$$ we have $|a_{n_k} - a_{n_j}| \le B_{N(\epsilon)} - A_{N(\epsilon)}=2^{-{N(\epsilon)}}(B-A) < \epsilon,$ which means that the subsequence $$(a_{n_k})_{k\in\mathbb N}$$ is a Cauchy sequence. By the completeness principle, it is convergent in $$\mathbb R$$.

Therefore, the limit of this convergent sequence is an accumulation point of the sequence.

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983