Show $f(x)=\sqrt{x}$ is continuous on $[0,1]$
For this question, I tried two ways to do it.
Suppose that sequence $\{x_n\}$ converges to $x_0\in[0,1]$ and $x_n\in[0,1]$ for all $n$. Then we have:
$$\lim\limits_{n\rightarrow\infty}f(x_n)=\lim\limits_{n\rightarrow\infty}\sqrt{x_n}=\sqrt{x_0}=f(x_0)$$
Therefore, $f(x)$ is continuous on $[0,1]$.
Let $\epsilon>0$ and $x_0\in[0,1]$, then there exists a $\delta=2\epsilon$ such that $|x-x_0|<\delta$ with for all $x\in[0,1]$. And for all $x,x_0\in[0,1]$, $\sup\{\sqrt{x}+\sqrt{x_0}\}=2$. Then:
$$|f(x)-f(x_0)|=|\sqrt{x}-\sqrt{x_0}|=\left|\frac{x-x_0}{\sqrt{x}+\sqrt{x_0}}\right|\leq|x-x_0|\left|\frac{1}{\sqrt{x}+\sqrt{x_0}}\right|<\delta/2=\epsilon$$
Thus, $f(x):[0,1]\rightarrow\mathbb{R}$ is continuous.
can someone check these two solution right or not? Thanks
Best Answer
Theorem: If $f$ is continuous and one-to-one mapping of compact metric space $X$ onto metric space $Y$. Then inverse mapping $f^{-1}$ defined $f^{-1}(f(x))=x$ is also continuous.
It's easy to see that $f(x)=x^2$ implies to all condition of above theorem. $X=[0,1]$ is compact. Hence $f^{-1}=\sqrt{x}$ is continuous.