Let $f(x)=x^2$. Prove that for any $E\subseteq\mathbb{R}$ the dimension of image is not changed i.e $$\dim_HE=\dim_H(f(E))$$
Any set in $\mathbb{R}$ can be represnted as a countable union of closed intervals e.g an open interval can be represented as $E=\bigcup_{n\in\mathbb{N}}[a+\frac{1}{n},b-\frac{1}{n}]$. So the dimension of $E$ equals to the dimension of the "longest" dimension in the union.
My problem is that I don't know how to prove that the dimension of closed interval is not changed under this function. I could have used $f(x)$ as a Lipschitz function but it gives only a bound. How can I prove it?