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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?

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  • 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!

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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.

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