A pasting lemma counterexample (of sorts)

[dkotschessaa]

From the pasting lemma we have that if ## X = A \cup B ## and ##f: A \rightarrow Y ## and ## g: A \rightarrow Y## are continuous functions that coincide on ## A \cup B ##, they combine to give a continuous function ## h: X \rightarrow Y ## s.t. ## h(x) = f(x) ## for ## x \in A ## and ## h(x) = f(g) ## for ## x \in B ##, so long as both ## A ## and ## B## are both open or closed.

But what if ##A## is open and ##B## is closed? This should not work. My reasoning is that, if it worked, then the pasting lemma would already say so! But this obviously does not constitute a proof or counterexample. I think it shouldn't be difficult, but I am missing something. The "obvious" choice was to take ##A## and ##B## as complements, but I am not sure how to show a lack of continuity. The intersection is empty, so the functions agree on that. Another example?

I feel like this should be very easy or obvious. A nudge in the right direction would be appreciated.

-Dave K

 

[fresh_42]

With complementary sets ##A,B## one could define ##f: (-\infty,0) \rightarrow \mathbb{R}## by ##1/x## and ##g: [0,+\infty) \rightarrow \mathbb{R}## by ##g(x)=0##. It would be more interesting / challenging to find an example with ##A \cap B \neq \emptyset##.

 

[dkotschessaa]

With complementary sets ##A,B## one could define ##f: (-\infty,0) \rightarrow \mathbb{R}## by ##1/x## and ##g: [0,+\infty) \rightarrow \mathbb{R}## by ##g(x)=0##. It would be more interesting / challenging to find an example with ##A \cap B \neq \emptyset##.

Indeed it would. This works for the time being or at least gets me started. I think my mental block was that the functions I was coming up with were all onto their domain even though there is no good reason for this. Thanks.

 

[fresh_42]

In case you find an example with overlapping domains, please post it. I'm curious now. (And confident, that there are some - I mean it's topology, isn't it?)

 

[dkotschessaa]

In case you find an example with overlapping domains, please post it. I'm curious now. (And confident, that there are some - I mean it's topology, isn't it?)

Will do.

 

[Infrared]

In case you find an example with overlapping domains, please post it. I'm curious now. (And confident, that there are some - I mean it's topology, isn't it?)

You can overlap the domains away from the problem points. Let [itex]X=S^1[/itex]. The usual angle function [itex]S^1\to [0,2\pi)[/itex] is of course discontinuous. But it continuous when restricted to the arc consisting of points with angles in [itex][0,\pi][/itex] and also to the arc of points with angles in [itex](\pi/2,2\pi)[/itex] (notice that the first arc is a closed subset of [itex]S^1[/itex] and the second is a open).

Edit: One could also have just enlarged the domain of your functions, so that, for example, [itex]f[/itex] is also defined on [itex](0,1)[/itex] and is zero here. But my example still holds if you want [itex]A[/itex] and [itex]B[/itex] to be connected.

 

[Mayaka Ibara]

In case you find an example with overlapping domains, please post it. I'm curious now. (And confident, that there are some - I mean it's topology, isn't it?)

Consider the functions ##f : (-\infty, 0]\cup\{1\} \rightarrow \mathbb{R} ##, defined by ##f(x)=1## and ##g: (0,\infty) \rightarrow \mathbb{R}##, defined by ##g(x)=x##.

 

[dkotschessaa]

Hey, I hadn't realized there were new posts on this thread. Thank you @Infrared and @Mayaka Ibara . I will check them out later.

 

[Bashyboy]

Consider the functions f:(−∞,0]∪{1}→Rf:(−∞,0]∪{1}→Rf : (-\infty, 0]\cup\{1\} \rightarrow \mathbb{R} , defined by f(x)=1f(x)=1f(x)=1 and g:(0,∞)→Rg:(0,∞)→Rg: (0,\infty) \rightarrow \mathbb{R}, defined by g(x)=xg(x)=xg(x)=x.

Is this actually a counterexample to the modified pasting lemma, where one set is suppose to be open and the other closed? The set ##(0, \infty)## is open in ##\mathbb{R}## endowed with the standard topology, but ##(-\infty, 0] \cup \{1\}## doesn't seem closed to me...I must be missing something.

 

[dkotschessaa]

Is this actually a counterexample to the modified pasting lemma, where one set is suppose to be open and the other closed? The set ##(0, \infty)## is open in ##\mathbb{R}## endowed with the standard topology, but ##(-\infty, 0] \cup \{1\}## doesn't seem closed to me...I must be missing something.

##(-\infty, 0] ## is closed since it's complement is ##(0, \infty)## which is open. The single point set ##\{1\}## is closed. (My reasoning - I know ##\mathbb{R}## is Hausdorff, implying it's ##T_1## so single point sets are closed. There might be an easier explanation.) So it's a union of closed sets.

Heaven help me if I'm wrong. My qualifier is in less than 60 days. :)

-Dave K