# Proof of the Intermediate value theorem using convergent sequences.

Theorem: If $$f$$ is continuous on the interval $$[a,b]$$ then $$\forall y \in [f(a),f(b)], \space \exists c \in [a,b]$$ such that $$f(c)=y$$

I've seen many proofs using the epsilon-delta definition of continuous maps. So I wanted to prove it using a slightly different approach using the definition of continuous maps as maps that preserve sequential convergence.

Proof attempt:

Define $$A:=\{x \in [a,b] : f(x)

Claim: if $$c=\sup A$$, then $$f(c)=y$$

We show it by eliminating other possibilities. Suppose $$f(c). Then let $$(x_{n})$$ be a sequence with $$(x_{n}) \to c$$ and $$x_{n} >c$$ for all $$n \in \mathbb{N}$$. The continuity of $$f$$ preserves the sequenial convergence of $$(x_{n})$$ so $$f(x_{n}) \to f(c)$$. If $$(f(x_{n_{k}}))$$ is a monotonic subsequence of $$(f(x_{n}))$$, then we know that there exists $$N \in \mathbb{N}$$ such that $$\forall n^*\geq N$$ where $$n^* \in \{n_{k}\}$$, we have $$f(c). But then $$x_{n^*} \in A$$ but $$c= \sup A for all $$n \in \mathbb{N}$$. So $$f(c)\geq y$$.

On the other hand, if $$f(c)>y$$, we can similarly construct a sequence with $$(x_{n}) \to c$$ but $$x_{n} \in A$$ for all $$n \in \mathbb{N}$$. We again find $$N \in \mathbb{N}$$ such that $$\forall n^*\geq N$$ we have $$y. But then $$x_{n^*}$$ is an upper bound which impossible since $$x_{n} for all $$n \in \mathbb{N}$$.

We can conclude $$f(c)=y$$

• If $x_n > c$, why $f(x_{n^*}) < y$? If $x_n >c$, then $x_n \notin A$, so $f(x_n)\geqslant y$. Taking the limit, $f(c) \geqslant y$. – xbh Dec 6 '18 at 14:06
• Alright. My bad. Your reasoning is acceptable. However I would only get $f(x_n) < y$ after some $N$. Why $f(c)<f(x_n)$ as well? – xbh Dec 6 '18 at 14:41
• So the sequence $(f(x_{n})) \to f(c)$. We can construct a monotonic subsequence of $(f(x_{n}))$ then either $f(c)<f(x_{n_{k}})$ or $f(x_{n_{k}})<f(c)$. If $f(x_{n_{k}})<f(c)$ then $f(x_{n_{k}})<f(c)<y$ but then $x_{n_{k}} \in A$ so $f(c)<f(x_{n_{k}})$. But the assumption is that $f(c)<y$ so we can find $f(x_{n_{k}^*})$ such that $f(c)<f(x_{n_{k}^*})<y$. – Sei Sakata Dec 6 '18 at 14:44
• I hope its clearer now. – Sei Sakata Dec 6 '18 at 14:47
• Okay. That is not that obvious, so it might be better if you include these in your final proof. Also I think my reasoning would be quicker. Additionally, for the 2nd case I think your $(x_n)$ should be picked from $A$. $x_n <c$ is clearly not sufficient. – xbh Dec 6 '18 at 14:52