# Questions about the proof of the intermediate value theorem

Let$$f:[a,b]\rightarrow \mathbb{R}$$ be continious. Then for every $$y_0\in[\min(f(a),f(b)), \max (f(a),f(b))]$$ there exists at least one $$x_0\in[a,b]$$ with $$f(x_0)=y_0$$.

Proof: Without loss of generality let $$f(a)\leq f(b)$$.

We define the set $$M$$

$$M:=\{x\in[a,b]:f(x)\geq y_0\}=f^{-1}([y_0,+\infty))\subseteq [a,b].$$

Then $$b\in M$$, thus $$M \neq \emptyset$$. Let $$x_0 := \inf M \in [a,b]$$. There exists a sequence $$(x_n)$$ in $$M$$ with $$x_n\rightarrow x_0$$. Because $$f$$ is continious, we have $$f(x_n)\rightarrow f(x_0)$$, and because $$f(x_n)\geq y_0$$ we also have $$f(x_0)\geq y_0$$, in conclusion $$x_0\in M$$, thus $$x_0= \min M$$.

I have understood the rest of the proof but if somebody wants me to write it down I will do it.

For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $$f(x_n)\geq y_0$$ we also have $$f(x_0)\geq y_0$$..." Every element of the sequence is $$\geq y_0$$, why must the same also apply to the Limit?

• Since $$M \subset [a,b]$$ it follows that $$a$$ is a lower bound.
• For the same reason you have $$a\leq \inf M$$.
• In general it holds $$c \leq y_n \stackrel{n\to\infty}{\rightarrow} y_0 \Rightarrow c \leq y_0$$. You can see this quickly by contradiction:
Assume $$y_0 < c$$. Choose $$\epsilon > 0$$ with $$y_0 + \epsilon < c \stackrel{y_n \stackrel{n\to\infty}{\rightarrow} y_0}{\Longrightarrow} c \leq y_n < y_0 + \epsilon < c$$ for sufficient large $$n$$. Contradiction!
The infimum of a set lies in the closure of the set (see here). So $$\inf M$$ lies in $$\overline M \subseteq \overline {[a, b]} = [a, b]$$. Basically, the infimum is in $$[a, b]$$ because it is a closed interval.
For your other question, yes: if all elements of a converging sequence are $$\geq y_0$$, then the limit is also $$\geq y_0$$. The same thing holds for "$$\leq$$". You can prove it by contradiction: assume the limit is $$\lneqq y$$. So you have $$d := y_0-f(x_0) \gneqq 0$$. By definition of convergence, there exists $$N$$ sufficiently big such that $$f(x_n)-f(x_0) \leq \frac d2$$, therefore $$y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) \gneqq d - \frac d2 \gneqq 0$$ for $$n > N$$ and this is a contradiction.