proof of spaces homeomorphic to Baire space
We choose subsets of for integers and satisfying the following.
This can be done inductively. Suppose that has already been chosen. As it is open, condition 3 says that it is not compact. Therefore, there is a such that has no finite open cover consisting of sets of diameter no more than (see here (http://planetmath.org/ProofThatAMetricSpaceIsCompactIfAndOnlyIfItIsCompleteAndTotallyBounded)). However, as Polish spaces are separable, there is a countable sequence of open sets with diameter less than and covering . As the space is zero dimensional, these can be taken to be clopen. By replacing by we can assume that . Then, replacing by , the sets can be taken to be pairwise disjoint.
satisfies the required properties.
We now define a function such that for each and . Choose any there is a sequence . This set has diameter bounded by and, so, for . This sequence is Cauchy (http://planetmath.org/CauchySequence) and, by completeness of the metric, must converge to a limit . As is closed, it contains for each and therefore
In fact, as it has zero diameter, this set must contain a single element, which we define to be .
If then, letting be the first integer for which , the sets and are disjoint and, therefore, and is one to one.
Finally, we show that is an open map, so that its inverse is continuous. Sets of the form
form a basis for the topology on . Then, is open and, therefore, is an open map.
|Title||proof of spaces homeomorphic to Baire space|
|Date of creation||2013-03-22 18:46:51|
|Last modified on||2013-03-22 18:46:51|
|Last modified by||gel (22282)|