# Hausdorff dimension from above

Let $$A_n$$ be a set of Hausdorff dimension $$1-\frac{1}{n}$$ then, the set $$A=\cup_n A_n$$ Has Hausdorff dimension $$1$$ (nevertheless having $$H_1(A)=0$$).

My question is: can we do the same thing from above? Given $$E_n$$ such that $$E_{n+1}\subset E_n$$ and $$dim_H(E_n)=1+\frac{1}{n}$$ is it true that $$dim_H(\cap_n E_n)=1?$$

For every $$s>1$$, there is $$n$$ with $$1+\frac1n and hence $$H^s(\bigcap E_n)\le H^s(E_n)=0$$.
However, it may happen that $$\dim_H(\bigcap E_n)<1$$. In fact, here's an example: Let Let $$A_n$$ be a compact set of Hausdorff dimension $$1+\frac1n$$ and such that the $$A_n$$ are disjoint and getting smaller so that $$E_n=\{0\}\cup \bigcup_{k\ge n} A_n$$ is still bounded. This makes $$\bigcap E_n=\{0\}$$