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

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