Proof: By Induction

(related to Theorem: Every Bounded Real Sequence has a Convergent Subsequence)

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