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It is my first post, so please be indulgent!

Here is the problem: I am in the class S of closed subsets of [0,1]^2 that are connected and have perimeter less or equal to 1.

I endow this space with the Hausdorff metric (or equivalently Fell topology), that says that for two compacts A, B, d(A,B)<=r iff every point of A is at distance less than r of B, and the other way around.

Question: Is this space compact?

Equivalent question: Is the function "perimeter" lower semi continuous on this set? (It is equivalent because the Hausdorff metric is compact for the class of all compact sets, and the class of connected sets is closed).

In other words, given a sequence of connected sets with perimeter at most 1, is it possible to see suddenly some perimeter appear at the limit? (which would be an answer "no" to the question).

The perimeter here is the 1-dimensional measure of the boundary, i.e. the infimum over all coverings of the boundary by balls of the sum of the diameters of the balls.

Remark: If one drops the assumption of connectedness, it is not true anymore (consider for instance a set of points -thus with zero perimeter- that becomes dense in the square). So this assumption is very important!

Thak you in advance!

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Hi Remark: I don't know the answer to your question, but here's a potential strategy. As Tom Leinster has been explaining for some time now (see eg his posts on n-Category Cafe), another good definition of perimeter of a closed set $X$ is $\frac{\partial}{\partial\epsilon} \bigr|_{\epsilon=0} \mathrm{Vol}(X_\epsilon)$, where $X_\epsilon$ is the set of all points within distance ϵ of X. Perhaps this is the same as your definition. Anyway, then perhaps you can successfully swap a limit somewhere when computing a limit in Hausdorff topology. – Theo Johnson-Freyd Aug 3 '11 at 19:06
up vote 12 down vote accepted

No. Let $B$ be a closed ball of radius 1/2 and $I$ a diameter of $B$. Construct a Cantor-like set $K\subset I$ of lengths $\mathcal H^1(K)=0.9$ (where $\mathcal H^1$ denotes the 1-dimensional Hausdorff measure). We have $I\setminus K=\bigcup I_i$ where $I_1,I_2,\dots$ are disjoint open subintervals of $I$ and $\sum_{i=1}^\infty\mathcal H_1(I_i)=0.1$. Let $B_i$ be the open ball for which $I_i$ is a diameter, and let $X_k=B\setminus\bigcup_{i=1}^k B_i$. Then each $X_k$ is a closed connected set and the sequence $\{X_k\}$ Hausdorff converges to $X=\bigcap_{k=1}^\infty X_k=B\setminus\bigcup_{i=1}^\infty B_i$.

The boundary of $X_k$ is a union of $k+1$ disjoint circles $\partial B$ and $\partial B_i$, $1\le i\le k$, hence $$ \mathcal H^1(\partial X_k) = \mathcal H^1(\partial B)+\sum_{i=1}^k \mathcal H^1(\partial B_i) = \pi+\sum_{i=1}^k\pi\mathcal H^1(I_i) \le \pi+0.1\pi =: C. $$ However the boundary of $X$ contains $K$, hence $\mathcal H^1(\partial X)\ge\mathcal H^1(\partial B)+\mathcal H^1(K)=\pi+0.9> C$.

Thus the space of connected sets of perimeter $\le C$ is not closed.

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Nice example. It might be interesting to notice that when you consider a subset of the boundary with the Cantor set is removed, the result is true. See R. Cerf, The Hausdorff Lower Semicontinuous Envelope Of The Length In The Plane Ann. Scuola Norm. Sup. Pisa Cl. Sci. (5) I (2002), 33$-$71.… – Tapio Rajala Aug 4 '11 at 9:51
Very nice example (I actually believed the answer would be yes), Cantor sets are even more vicious than I thought. Thank you very much! – kaleidoscop Aug 5 '11 at 8:25

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