Pasting Lemma
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From: William Elliot on 8 Jun 2010 03:35 On Mon, 7 Jun 2010, Marc Olschok wrote: > William Elliot <marsh(a)rdrop.remove.com> wrote: Hi Marc. >> Does the following form of the pasting lemma hold for all spaces? >> >> If D is dense, K closed, U open, X = K \/ U, f:X -> Y >> and f is continuous over U and over K \/ D, then f is >> continuous. > > Here is a small counterexample: > > Consider X = { a , b , c , d } with three > nontrivial open sets: { a , b }, { b , d } , { b }. > and { a,b,d }. {c} is closed. X is not regular. > Let K = { c , d } , U = { a , b } and D = { b , d } > and define f: X --> X by > f(a) = a and f(x) = b for all x =/= a > > Then the restriction of f to K \/ D is constant, hence continuous > and the restriction of f to U is just the inclusion of the open > subspace U and therefore also continuous. Yes, f|U is the identity map. > But the preimage of the closed subset { a , c } under f is { a }, > which is not closed. > Most excellent. At Ask-a-Topologist, http:at.yorku.ca/topology in the topology forum there is a discussion going on about net continuity on a dense set. Indications are that the pasting lemma is a corollary to the more general proposition under discussion, that requires Y to be regular. With your reminder, I'm prompted to use my favorite irregular space. Let Rt be R with the topology generated by the open intervals and Q. Rt is irreglar. Ok, I get a different counterexample: Rt; D = Q = U, K = {0}; f:Rt -> Rt, f(Q) = {0}, f(R\Q) = {1}. Oh oh. The same counter example holds if f:Rt -> R and R is regular. Shucks, the pasting lemma isn't a corollary to the net continuity proposition. -- Don't buy BP gas. They caused the gulf mess by dodging safety requirements, government regulation and reckless cost cutting. ----
From: William Elliot on 8 Jun 2010 07:28
On Tue, 8 Jun 2010, William Elliot wrote: > On Mon, 7 Jun 2010, Marc Olschok wrote: >> William Elliot <marsh(a)rdrop.remove.com> wrote: > > Hi Marc. > >>> Does the following form of the pasting lemma hold for all spaces? >>> >>> If D is dense, K closed, U open, X = K \/ U, f:X -> Y >>> and f is continuous over U and over K \/ D, then f is >>> continuous. >> >> Here is a small counterexample: >> >> Consider X = { a , b , c , d } with three >> nontrivial open sets: { a , b }, { b , d } , { b }. >> > and { a,b,d }. {c} is closed. X is not regular. > >> Let K = { c , d } , U = { a , b } and D = { b , d } >> and define f: X --> X by >> f(a) = a and f(x) = b for all x =/= a >> >> Then the restriction of f to K \/ D is constant, hence continuous >> and the restriction of f to U is just the inclusion of the open >> subspace U and therefore also continuous. > > Yes, f|U is the identity map. > >> But the preimage of the closed subset { a , c } under f is { a }, >> which is not closed. >> > Most excellent. At Ask-a-Topologist, http:at.yorku.ca/topology > in the topology forum there is a discussion going on about > net continuity on a dense set. Indications are that the > pasting lemma is a corollary to the more general proposition > under discussion, that requires Y to be regular. > > With your reminder, I'm prompted to use my favorite irregular space. > Let Rt be R with the topology generated by the open intervals and Q. > Rt is irreglar. Ok, I get a different counterexample: > Rt; D = Q = U, K = {0}; f:Rt -> Rt, f(Q) = {0}, f(R\Q) = {1}. > > Oh oh. The same counter example holds if f:Rt -> R > and R is regular. Shucks, the pasting lemma isn't > a corollary to the net continuity proposition. > Well hurrah, it's not a counter example. I ommited checking for X = K \/ U. The new prefered version is dense D, open U, regular Y, f:X -> Y, f continuous over U and over X\U \/ D implies f is continuous. > -- > Don't buy BP gas. They caused the gulf mess by dodging > safety requirements, government regulation and reckless > cost cutting. > > ---- > |