# Is the pasting lemma also true in case of uniform continuity?

Let $\left(X, d_X \right)$ and $\left( Y, d_Y \right)$ be metric spaces; let $A$ and $B$ be (non-empty) closed subsets of $X$ such that $X = A \cup B$; let $f \colon A \to Y$ and $g \colon B \to Y$ be uniformly continuous functions such that $f(x) = g(x)$ for all $x \in A \cap B$; and, let the function $h \colon X \to Y$ be defined as follows: $$h(x) \colon= \begin{cases} f(x) \ & \mbox{ if } \ x \in A, \\ g(x) \ & \mbox{ if } \ x \in B. \end{cases}$$ Then is $h$ also uniformly continuous on $X$?

I know that $h$ is continuous, even if both $f$ and $g$ were merely continuous, which is Theorem 18.3 (The Pasting Lemma) in the book Topology by James R. Munkres, 2nd edition.

My Attempt:

Let $\varepsilon > 0$ be given. We need to find a real number $\delta > 0$ such that $$d_Y \left( h \left( x \right) , h \left( x^\prime \right) \right) < \varepsilon$$ for every pair of points $x, x^\prime \in X$ for which $$d_X \left( x, x^\prime \right) < \delta.$$

Now if $x, x^\prime \in A$ or if $x, x^\prime \in B$, then this is of course possible.

What if $x \in A \setminus B$ and $x^\prime \in B \setminus A$?

• The answer given by Aloizio Macedo is the exact reason why this can fail, but it would be interesting to ask if the statement is true with an additional assumption, that every ball is connected. I believe it will be true. – Adayah Nov 5 '17 at 13:42
• @Adayah thank you for your comment. Can you please state and prove the result you have in mind? – Saaqib Mahmood Nov 5 '17 at 13:47
• Karina Livramento did that before me. ;-) – Adayah Nov 5 '17 at 16:08

The case $X = \mathbb{R}^n$ is true. More generally, whenever $X$ has the property that every ball is path connected, this Pasting Lemma holds. Indeed, given $\epsilon > 0$, there exists $\delta > 0$ such that

$$x, y \in A \ \ \text{and}\ \ d(x, y) < \delta \implies d(f(x), f(y)) < \frac{\epsilon}{2}\\ x, y \in B \ \ \text{and}\ \ d(x, y) < \delta \implies d(g(x), g(y)) < \frac{\epsilon}{2}$$

Then, if $x \in A$ and $y \in B$, with $d(x, y) < \delta/2$, we can choose a point $z \in B(x; \delta/2)$ that is in $A \cap B$ (consider the path connecting $x$ and $y$ inside the ball). We have

$$d(x, z) <\delta, \quad d(z, y) < \delta$$ Therefore $$d(h(x), h(y)) \leq d(f(x), f(z)) + d(g(z), g(y)) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

The uniform continuity follows.

• Path-connectedness of balls can be relaxed to connectedness. Since $B(x, \delta/2)$ is connected and is covered by two closed sets $A, B$, there must be some $z \in A \cap B$ in the ball. – Adayah Nov 5 '17 at 16:03
• For the sake of explicitness for the OP, the property that "every ball is path-connected/connected" cannot be relaxed to the more well-known property of local path-connectedness/connectedness. The counter-example in my answer is locally path-connected. – Aloizio Macedo Nov 5 '17 at 16:12

No, it is not true.

Take $A=\{(x,1/x) \mid x>0\} \subset \mathbb{R}^2$, $B=\{(x,0) \mid x>0)\} \subset \mathbb{R}^2,$ $X:=A \cup B$. Both $A,B$ are closed in $X$, with their intersection being empty.

Now take $f:A \to \mathbb{R}$ given by $f((x,1/x))=x$, and $g:B \to \mathbb{R}$ given by $g((x,0))=2x$. Both are uniformly continuous, and satisfy the hypothesis of the alleged "pasting lemma" (coincide in the intersection). However, $h:A \cup B \to \mathbb{R}$ is not uniformly continuous. There are points arbitrarily close when you go further to the right in $A$ and $B$ and such that their images are arbitrarily distant.

• thanks, but can we also have an example where the pieces $f$ and $g$ of $h$ actually do coincide? Can we find an example of a function whose domain is the full space $\mathbb{R}$ or $\mathbb{R}^2$? – Saaqib Mahmood Nov 5 '17 at 13:44
• @SaaqibMahmuud It is easy to adapt to a situation where there is intersection. Just make a vertical line $c$ crossing $A,B$ somewhere, let $A:=A \cup c$, $B:=B \cup c$ and extend the functions linearly on $c$ such that they match to what they should. About the case where the domain of $h$ is the full $\mathbb{R}^2$, I don't know the answer right now. – Aloizio Macedo Nov 5 '17 at 13:48