Exercise: Let $(a_n)_{n=1}^\infty$ be a Cauchy sequence. Prove that $(a_n)_{n=1}^\infty$ is bounded.

Proof: Given that $(a_n)_{n=1}^\infty$ is a Cauchy sequence. For $\varepsilon=1$ there exists an $N>1$ such that for all $j,k>N$ we have that $|a_j-a_k|<1$. Thus, we split the sequence into a finite sequence $(a_n)_{n=1}^N$ and the remaining infinite sequence $(a_n)_{n=N+1}^\infty$

The finite sequence is bounded by some number $M>0$. For the remaining infinite sequence, we have the inequality $|a_j-a_k|<1$ for $j,k>N$. Adding $|a_k|$ to both sides we get $|a_j-a_k|+|a_k|<1+|a_k|$. Using the triangle inequality we get $|a_j-a_k+a_k|<1+|a_k|\implies |a_j|<1+|a_k|$. Since the finite sequence is bounded by $M$, the whole sequence $(a_n)_{n=1}^\infty$ is bounded by $1+|a_k|+M$.

Is the proof correct?


1 Answer 1


Not quite. Mathematically there are two things you should fix. Firstly, you need to explain where $M$ has come from. This should be easy, you can write down an $M$ that works.

Secondly though, you cannot leave $|a_k|$ in the final upper bound because - and this is slightly weird to understand - you never really 'fixed' $k$. (When you introduced the index $k$ you basically just had $\forall\ j,k > N$)

  • $\begingroup$ Thanks for that! $\endgroup$
    – Seeker
    Commented Aug 13, 2022 at 23:29

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