Pasting Lemma
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From: William Elliot on 1 Jun 2010 02:28 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. It is a result of the following which is known to hold for metric spaces. If D is dense, U open and dense, A = bd U \/ U/\D, f:X -> Y, and f is continuous over U and over A, then f is continuous.
From: José Carlos Santos on 1 Jun 2010 03:50 On 01-06-2010 7:28, William Elliot wrote: > 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. Not true. Take K = empty set. Then what you are saying is: if _f_ is a function from X into Y which is continuous over a dense set, then _f_ is continuous. I suppose that you will be able to find counter-examples easily. Best regards, Jose Carlos Santos
From: Ron on 1 Jun 2010 10:10 On Jun 1, 3:50 am, José Carlos Santos <jcsan...(a)fc.up.pt> wrote: > On 01-06-2010 7:28, William Elliot wrote: > > > 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. > > Not true. Take K = empty set. Then what you are saying is: if _f_ is a > function from X into Y which is continuous over a dense set, then _f_ is > continuous. I suppose that you will be able to find counter-examples easily. > Not quite. the OP stated "continuous over U and over K \/ D". If K is the empty set then U=X, so in this case his statement would be "If f is continuous over X and over D, it is continuous" which is trivially true. > Best regards, > > Jose Carlos Santos
From: José Carlos Santos on 1 Jun 2010 12:23 On 01-06-2010 15:10, Ron wrote: >>> 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. >> >> Not true. Take K = empty set. Then what you are saying is: if _f_ is a >> function from X into Y which is continuous over a dense set, then _f_ is >> continuous. I suppose that you will be able to find counter-examples easily. >> > > Not quite. the OP stated "continuous over U and over K \/ D". If K is > the empty set then U=X, so in this case his statement would be "If f > is continuous over X and over D, it is continuous" which is trivially > true. Right. My mistake. :-( Best regards, Jose Carlos Santos
From: Marc Olschok on 7 Jun 2010 14:00
William Elliot <marsh(a)rdrop.remove.com> wrote: > 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 }. 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. But the preimage of the closed subset { a , c } under f is { a }, which is not closed. -- Marc |