Real Continuous Functions

Def Continuity A function $f:X\to Y$ on $\mathbb{R}$ is continuous at $x_0$ if for every $\{a_n\}\subset X$ such that $a_n\to x_0$, we have $f(a_n)\to f(x_0)$. Equivalently, it is continuous at $x_0$ if for all $\epsilon >0$ there exists $\delta >0$ such that if $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$. We say $f$ is continuous on $X$ if it is continuous at every $x_0\in X$. e.g. Polynomial functions are continuous on $\mathbb{R}$.

Continuous Functions Theorems

Thrm If $f:[a,b]\to \mathbb{R}$ is continuous, then $f$ is bounded, and attains its bounds.

Thrm Intermediate Value Theorem If $f:[a,b]\to \mathbb{R}$ is continuous, then for any real $\gamma$ such that $\min (f(a),f(b))\le \gamma \le \max (f(a),f(b))$, there exists $c\in [a,b]$ such that $f(c)=\gamma$. Proof Without loss of generality, assume $f(a)<f(b)$. Let $A=\{x\in [a,b]:f(x)\le \gamma \}$. As $A\subset [a,b]$, $A$ is bounded. Let $M=\sup (A)$. By lemma, there exists a sequence $\{a_n\}\subset A$ such that $a_n\to M$. As for all $n$, $f(a_n)\le \gamma$, we have $f(M)\le \gamma$. So $M\in A$. Now because