# A pasting lemma counterexample (of sorts)

• A
• dkotschessaa
In summary, the modified pasting lemma states that if ##X = A \cup B## and ##f: A \rightarrow Y## and ##g: B \rightarrow Y## are continuous functions that coincide on ##A \cap B##, they combine to give a continuous function ##h: X \rightarrow Y## such that ##h(x) = f(x)## for ##x \in A## and ##h(x) = g(x)## for ##x \in B##, as long as one of the sets ##A## or ##B## is open and the other is closed. Counterexamples have been provided to show that this lemma does not hold when both ##A## and ##B## are open or closed
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

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
fresh_42 said:
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.

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
fresh_42 said:
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.

fresh_42 said:
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.

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fresh_42 said:
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##.

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

Mayaka Ibara said:
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.

Bashyboy said:
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

Bashyboy

## 1. What is the "pasting lemma counterexample (of sorts)"?

The pasting lemma counterexample (of sorts) is a mathematical concept that provides a counterexample to the pasting lemma, which is a theorem in topology. It shows that under certain conditions, the pasting lemma does not hold.

## 2. What is the pasting lemma?

The pasting lemma is a theorem in topology that states that the union of two closed sets is also a closed set, as long as the two sets have a point in common on their boundary.

## 3. What are the conditions for the pasting lemma to hold?

The pasting lemma holds when the two closed sets have a point in common on their boundary. In other words, the boundary of one set must contain a point from the other set, and vice versa.

## 4. How does the pasting lemma counterexample (of sorts) work?

The pasting lemma counterexample (of sorts) works by providing a scenario where the conditions for the pasting lemma do not hold. It involves two closed sets with a point in common on their boundary, but the union of the two sets is not closed.

## 5. What are the implications of the pasting lemma counterexample (of sorts) in topology?

The pasting lemma counterexample (of sorts) highlights the importance of carefully considering the conditions of a theorem, as a seemingly simple condition can have significant implications. It also serves as a reminder that counterexamples are important tools in mathematics for understanding the limitations of theorems and theories.

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