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

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

We show it by eliminating other possibilities. Suppose $f(c)<y$. 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 $f(c)<f(x_{n^*})<y$. But then $x_{n^*} \in A$ but $c= \sup A<x_{n}$ 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 $y<f(x_{n^*})<f(c)$. But then $x_{n^*}$ is an upper bound which impossible since $x_{n}<c$ for all $n \in \mathbb{N}$.

We can conclude $f(c)=y$

  • $\begingroup$ 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, $f(c) \geqslant y$. $\endgroup$ – xbh Dec 6 '18 at 14:06
  • $\begingroup$ 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? $\endgroup$ – xbh Dec 6 '18 at 14:41
  • $\begingroup$ 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 $f(c)<f(x_{n_{k}^*})<y$. $\endgroup$ – Sei Sakata Dec 6 '18 at 14:44
  • $\begingroup$ I hope its clearer now. $\endgroup$ – Sei Sakata Dec 6 '18 at 14:47
  • $\begingroup$ 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. $\endgroup$ – xbh Dec 6 '18 at 14:52

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