Timeline for Arbitrary union of meagre open sets
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2012 at 19:31 | comment | added | Gerald Edgar | Of course you need choice even to prove a countable union of meager sets is meager. | |
| Oct 10, 2012 at 12:44 | comment | added | Joel David Hamkins | There is also use of AC in both arguments just to pick the covers $C_n^i$, and this is not just countable AC or DC, since one needs to choose a cover of $U_i$ for each $i$. | |
| Oct 10, 2012 at 12:11 | comment | added | Joel David Hamkins | Jochen, good question, and I had had the same thought. It appears that Theo's maximality argument may need a little less choice than my argument, since to construct a maximal set of his sort it suffices to well-order the basic open sets, but in my argument, I need to well-order the family of open meager sets, which might be much larger. For example, one can find maximal sets without any AC when there are only countably many basic open sets (but in this case the main theorem is immediate by countable unions). | |
| Oct 10, 2012 at 6:03 | comment | added | Jochen Wengenroth | Since Joel's proof uses well-ordering and Theo's maximality one may ask whether the result itself depends on the axiom of choice. | |
| Oct 9, 2012 at 20:06 | vote | accept | Yvoz | ||
| Oct 9, 2012 at 18:22 | comment | added | Joel David Hamkins | Theo, thanks. Meanwhile, I also found a way to do it by enumerating the family in a well-ordered sequence. But perhaps this may amount to essentially the same thing as the maximal family argument... | |
| Oct 9, 2012 at 18:19 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Figured out the full case; added 22 characters in body
|
| Oct 9, 2012 at 18:07 | comment | added | Theo Buehler | The statement of the question is called the Banach category theorem (see e.g. Oxtoby, Measure and category, Theorem 16.1). The general case can be reduced to your claim by considering the union $V = \bigcup V_j$ of an arbitrary family of meager open sets and taking a maximal family of pairwise disjoint open sets $U_i$ such that each $U_i$ is contained in some $V_j$. Then it follows that $\overline{V} \setminus \bigcup U_i$ is meager (otherwise the family $U_i$ wouldn't have been maximal) and your claim does the rest. | |
| Oct 9, 2012 at 17:46 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 31 characters in body
|
| Oct 9, 2012 at 17:31 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |