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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 $$dim_H(\cap_n E_n)=1?$$

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For every $s>1$, there is $n$ with $1+\frac1n<s$ 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 $\bigcap E_n=\{0\}$

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