# Proving that every Cauchy sequence is bounded. [duplicate]

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?

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 $$\forall\ j,k > N$$)