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The interval $[0,1]$ is partitioned to $n$ disjoint parts. Is the set of all possible partitions compact?

There are several cases:

A. All $n$ parts are connected intervals (possibly empty). In this case, every partition can be represented as a vector with $n-1$ elements in $[0,1]$ (representing the locations of cuts). Hence the set of all partitions is equivalent to $[0,1]^{n-1}$, which is compact by the Tychonoff theorem.

B. The $n$ parts are arbitrary subsets of $[0,1]$. In this case, every partition can be represented as a function from $[0,1]$ to the finite set $\{1,...,n\}$, where $f$ assigns to each point the number of the part it belongs to. The set of such functions is $\{1,...,n\}^{[0,1]}$, which is also compact by the Tychonoff theorem.

C. The $n$ parts are measurable subsets of $[0,1]$. A partition can be represented as in B., but now $f$ has to be measurable. Is the subset of measurable functions compact in the Tychonoff topology of $\{1,...,n\}^{[0,1]}$?

Peter Taylor
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Erel Segal-Halevi
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  • I'm not sure how do you define a natural topology in the set of all partitions, care to explain? – PenasRaul Jun 19 '14 at 21:31
  • In cases A and B, the set of all partitions is equivalent (I think) to another set which has a well-known natural topology. In case C, I don't know... that is part of my question. – Erel Segal-Halevi Jun 19 '14 at 21:34
  • From my point of view you can make up some arbitrary topology and claim that is compact. I want to know that is the "connection" between the topologies you choose and the partition itself. – PenasRaul Jun 19 '14 at 21:36
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    Part B is more general than part C... Also, how do you represent $n$ arbitrary subsets with a function $f:{1,...,n}\to [0,1]$? Perhaps you mean $f:[0,1]\to {1,...,n}$, where the function "labels" a point $x$ in $[0,1]$ with the number of the set it belongs to? – Kyle Jun 19 '14 at 22:19
  • And then part C can be represented same as part B, only $f$ has to be measurable. – Conifold Jun 19 '14 at 22:26
  • @Conifold thanks a lot for editing my question. This is just what I meant! – Erel Segal-Halevi Jun 20 '14 at 05:20

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Measurable partitions are not compact. Take $n=2$ and let's use $0,1$ as labels instead of $1,2$. Then the labeling functions are exactly the characteristic functions of measurable subsets of $[0,1]$. For them the Tychonoff topology is just the topology of pointwise convergence, and it follows from the Tychonoff theorem that a subspace of $\{0,1\}^{[0,1]}$ is compact if and only if it is closed (Kelley, General Topology, Theorem 7.1).

Now let $A$ be a non-measurable subset of $[0,1]$, its finite subsets $F\subset A$ are directed by inclusion, and the net $\chi_F$ converges pointwise to $\chi_A$ because for any point all finite subsets containing the ones with it also contain it. But $\chi_F$ are measurable and $\chi_A$ is not.

So the subspace of measurable functions is not closed, hence not compact. This is not exactly surprising, with measurable functions any kind of closure can only involve sequences, and the Tychonoff topology on $\{0,1\}^{[0,1]}$ is not first countable. A pointwise limit of a sequence of measurable functions is always measurable, but I am not sure if this subspace is sequentially compact either, there is another problem, see What's going on with "compact implies sequentially compact"?

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Conifold
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