Proof if $f(x)$ is continuous on $[a,b]$,strictly increasing on $(a,b)$,then $f(x)$ is strictly increasing on $[a,b]$ My attempt:
Because $f(x)$ is continuous at $a$,then $\lim_{x\rightarrow a^+}f(x)=f(a)$
When $x\rightarrow a^+$,$f(x)$ is strictly decreasing,so for any $x \in (a,b)$,  $f(x)>f(a)$. It can also show that $f(x)<f(b)$ in the same way.
Do I prove it in a rigorous way?Thanks in advance!
 A: The idea is very simple. Since $f$ is strictly increasing on $(a, b)$ it follows that if we have $a < x < b$ then we can choose $y, z$ such that $a < y < z < x < b$ and then $f(y) < f(z) < f(x)$. Letting $y \to a^{+}$ we get $f(a) \leq f(z) < f(x)$ and thus $f(a) < f(x)$. Similarly it can be proved that $f(x) < f(b)$. It follows that $f$ is strictly increasing on $[a, b]$.
A: Since $f$ is continuous at $a$, then as you wrote $\lim_{x\rightarrow a^+}f(x)=f(a)$. Let $\{x_n\}$ be a strictly decreasing series, such that $\lim_{n\rightarrow \infty} x_n=a$. From $f's$ continuity $\lim_{n\rightarrow \infty} f(x_n)=f(a)$. Let's show that $f(a)<f(x_n)$ for all $n\in \mathbb{N}$. If exists $n_0$ such that $f(x_{n_0})\leq f(a)$, then for all $n\geq n_0$ we have that $f(x_n)<f(x_{n_0})\leq f(a)$. Since $f$ is continuous, then for all $\varepsilon >0$ exists $N\in \mathbb{N}$ such that for all $n\geq N$, $|f(x_n)-f(a)|<\varepsilon$. In particular, this is true for $\varepsilon=\frac{f(a)-f(x_{n_0})}{2}$. But for all $N\in \mathbb{N}$ we have some $n>\max\{n_0, N\}$ such that $|f(x_n)-f(a)|=f(a)-f(x_n)>f(a)-f(x_{n_0})>\frac{f(a)-f(x_{n_0}}{2}=\varepsilon$, contradicting continuity! Therefore, $f(a)< f(x_n)$ for all $x_n$, such that $f(a)<f(x)$ for all $a<x$.
Likewise, you can prove that $f(x)<f(b)$ for all $x<b$.
