Prove that a decreasing sequence of compact sets converges in the Hausdorff metric

Question

Let $A_n$ be a decreasing sequence of nonempty compact sets in $X$. Then I want to prove that:

$$\displaystyle \lim A_n = \bigcap_{n\ge 1} A_n$$

in $\mathcal C_H$. Where $$C_X=\{\text{non empty compact subsets of X}\}$$ and

$$d_H(A, B) = \inf\{ \epsilon > 0: A \subset U_{\epsilon}(B),\ \ B\subset U_\epsilon(A)\}.$$

So I wanted to adapt the proof given in page 11 of Kigami's book Analysis on fractals, then I did the following:

Let $D_2= \bigcap_{n\ge 1} A_n$ and we pick $x \in D_2$ then $x \in A_n$ so $d(x,A_n)<\epsilon$ because $d(x,A)=0$.

Now, there must exists a $N$ such that $U_{\epsilon}(D_2) \supset A_N$, because if this is not true then we can take for each $n$, $x_n \in A_n$ such that $d_H(x_n,D_2)>\epsilon$ and since each $A_n$ is compact and we have a decreasing sequence, then there exists a subsequence $x_{n_k} \to x_0$ and $x_0 \in D_2$ so this is a contradiction.

And we observe that for $n \ge N$ we have $A_n \subset A_N$, then $U_{\epsilon}(D_2) \supset A_n$, and $U_{\epsilon}(A_n) \supset D_2$, therfore we have:

$$d_H(A_n, D_2)<\epsilon$$

Is my adapted proof well performed? or how an I fix my mistakes?

Note: $U_{\epsilon}(A)=\{x\in X: d(x,A)<\epsilon\}$.

Answer 1

Your proof is correct. I have two suggestions for improvement:

• rename $D_2$ into $D$ or something else (the index is unnecessary and confusing)
• change the argument by contradiction to the following: the sets $A_n\setminus U_\epsilon (D)$ are compact, nested, and have empty intersection; therefore they are empty for sufficiently large $n$.