PomeAFBD46E7C0A4Theorem 19.7.6.AFBD46E7C0A4 Let $K_ i, K\in \mathcal{K}_ n$ ($i\in \mathbb {N}$). Then $K_ i\xrightarrow {d_{\mathrm{Haus}}}K$ if and only if the following conditions hold:
45F2C2B839F6each point $x\in K$ is the limit of a sequence $x_ i\in K_ i$, and
BA71889FAAC3the limit of any convergent sequence $(x_{i_ j})_{j\in \mathbb {N}}$ with $x_{i_ j}\in K_{i_ j}$ lies in $K$, where $i_ j$ is a strictly increasing sequence in $\mathbb {Z}_{>0}$.
Proof. Hausdorff convergence immediately supplies points $x_i\in K_i$ tending to each $x\in K$, and forces every convergent selection from a subsequence to land in $K$.
Conversely, the two conditions first imply that the $K_i$ are uniformly bounded. Otherwise choose a subsequence containing points $y_i$ tending to infinity, and, by the first condition, points $x_i\in K_i$ tending to a fixed $x\in K$. Choose $R$ with $K\subset B(x,R)$. Convexity gives $z_i\in[x_i,y_i]\subset K_i$ with $|z_i-x|=R$ for large $i$. A subsequence of these sphere points converges outside $K$, contradicting the second condition.
If the Hausdorff distance did not tend to zero, one of two failures would occur along a subsequence. Either there are $y_i\in K_i$ at a fixed positive distance from $K$; uniform boundedness gives a convergent subsequence, contradicting the second condition. Or there are $x_i\in K$ at a fixed positive distance from $K_i$; pass to $x_i\to x\in K$ and use the first condition to find $y_i\in K_i$ with $y_i\to x$, again a contradiction. This proves the converse and explains why convexity is needed in the boundedness step.
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.