If you have a decreasing sequence of compact subsets of X such that C1⊃C2⊃... Cm and C=∩∞,m=1, then show that the Hausdorff metric, D(Cn,C)→0 as n→∞.

Assume C∈S(X)

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- Feb 2nd 2010, 10:56 AMSamBourneHelp with fractals
If you have a decreasing sequence of compact subsets of X such that C1⊃C2⊃... Cm and C=∩∞,m=1, then show that the Hausdorff metric, D(Cn,C)→0 as n→∞.

Assume C∈S(X) - Feb 2nd 2010, 11:32 AMOpalg
Since C is the intersection of the $\displaystyle C_m$s, it is sufficient to find a value of m for which every element of $\displaystyle C_m$ is within distance $\displaystyle \varepsilon$ of C (for some given $\displaystyle \varepsilon>0$).

Let $\displaystyle C_\varepsilon = \{x\in C_1:d(x,C)\geqslant\varepsilon\}$. This is a closed (and therefore compact) subset of $\displaystyle C_1$. It is covered by the sets $\displaystyle U_n = \{x\in C_1:x\notin C_n\}\ (n\geqslant1)$, which are open in $\displaystyle C_1$. By compactness there is a finite subcover, and since the sets $\displaystyle U_n$ form an increasing nest, there is in fact just one of them, say $\displaystyle U_m$, that contains $\displaystyle C_\varepsilon$. It follows by taking complements that $\displaystyle C_m\subseteq C_\varepsilon$. Thus $\displaystyle d(C_m,C)<\varepsilon$.