The closure of the union equals the union of the closures
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From: Edward Green on 9 Jul 2010 17:47 The closure of the union equals the union of the closures: this is stated as a theorem to be proved shortly after defining the closure of a set, and I've been beating my head against the silly thing. Can anybody give me a hint in the right direction? Just how should I approach such a problem?
From: Daniel Giaimo on 9 Jul 2010 18:19 On 7/9/2010 5:47 PM, Edward Green wrote: > The closure of the union equals the union of the closures: this is > stated as a theorem to be proved shortly after defining the closure of > a set, and I've been beating my head against the silly thing. Can > anybody give me a hint in the right direction? Just how should I > approach such a problem? This is quite clearly false as stated. For an easy counterexample, take any topological space in which points are closed. Then every set is the union of its singleton subsets all of which are closed. -- Dan G
From: Gerry on 9 Jul 2010 19:13 On Jul 10, 7:47 am, Edward Green <spamspamsp...(a)netzero.com> wrote: > The closure of the union equals the union of the closures: this is > stated as a theorem to be proved shortly after defining the closure of > a set, and I've been beating my head against the silly thing. Can > anybody give me a hint in the right direction? Just how should I > approach such a problem? Suppose x is in the closure of the union. Then there's a sequence of points in the union converging to x. That sequence has a subsequence consisting entirely of points in one of the original sets. -- GM
From: christian.bau on 9 Jul 2010 19:25 On Jul 9, 10:47 pm, Edward Green <spamspamsp...(a)netzero.com> wrote: > The closure of the union equals the union of the closures: this is > stated as a theorem to be proved shortly after defining the closure of > a set, and I've been beating my head against the silly thing. Can > anybody give me a hint in the right direction? Just how should I > approach such a problem? How do you approach it: There is a family F of sets, and you are to prove that the closure C of the union of all sets in F is equal to the union U of the closures of all the sets in F. You need to prove C = U. So assume that x is in C and prove that it is in U. Then assume that x is in U and prove that it is in C. With both directions proved you proved C = U. BTW. I think it depends on whether F is finite or not. Consider sets of real numbers. Let Sn = { 1 / n } for all integers n >= 1. Is 0 in C? Is 0 in U?
From: Rob Johnson on 9 Jul 2010 19:26
In article <1472e7f6-d72d-43dc-a757-39be2609cdfd(a)g19g2000yqc.googlegroups.com>, Edward Green <spamspamspam3(a)netzero.com> wrote: >The closure of the union equals the union of the closures: this is >stated as a theorem to be proved shortly after defining the closure of >a set, and I've been beating my head against the silly thing. Can >anybody give me a hint in the right direction? Just how should I >approach such a problem? If you are talking about a finite union, then it is true. In this case, Gerry Myerson gives an excellent hint. If you are talking about an infinite union, then it is false. In this case, consider the union of all single rational points in the reals. The closure of the union is the reals, while the union of the closures is the rationals. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |