If $f:(A, d_A)\longrightarrow (Y, d^{'})$ and $g:(B, d_B)\longrightarrow (Y, d^{'})$ are uniformly continuous then $h$ is uniformly continuous?

Question

Let $(X, d)$ and $(Y, d^{'})$ be metric spaces, $A, B\subset X$ subspaces such that $X=A\cup B$. Suppose $$f:(A, d_A)\longrightarrow (Y, d^{'})\quad \textrm{and}\quad g:(B, d_B)\longrightarrow (Y, d^{'}),$$ are both uniformly continuous such that $f|_{A\cap B}=g|_{A\cap B}$. Is it true that $h:(X, d)\longrightarrow (Y, d^{'})$ given by $$h(x)=\left\{\begin{array}{ccl} f(x)&\textrm{se}&x\in A\\ g(x)&\textrm{se}&x\in B\end{array}\right.$$ is also uniformly continuous?

What about the case $B=X\setminus A$ with $A$ compact? Finally, what if I suppose $h$ is continuous?

Notation: $d_A$ and $d_B$ are the restriction of the metric $d$ to the subspaces $A$ and $B$, respectively.

Obs: Maybe additional hypothesis will be needed.

Answer 1

Without connectedness we have the following counter example. Let $A = (-\infty,0)$ and $B = (0,\infty)$. Let $f : A \to \Bbb{R}$ and $g : B \to \Bbb{R}$ be given by $f \equiv 0$ and $g \equiv 1$. Then $X = A\cup B = \Bbb{R} \backslash \{0\}$ and $$ h(x) = \begin{cases} 0 & x < 0 \\ 1 & x > 0 \end{cases} $$ which is not uniformly continuous.

Answer 2

Counterexample for the problem as written:

$A = (0,1)$, $B = (1,2)$. Take $$ f(x) = \begin{cases} 0 & x \in A\\ 1 & x \in B \end{cases} $$ Then $f$ is not uniformly continuous on $A \cup B$.

This will work, however, in the case that both $A$ and $B$ are compact.

Proof given compactness:

We know the following: for every $\epsilon > 0$, there is a $\delta_1 > 0$ such that for all $x,y \in A$, we have $d(x,y) < \delta_1 \implies d'(h(x),h(y)) < \epsilon$ and for all $x,y \in B$, we have $d(x,y) < \delta_1 \implies d'(h(x),h(y)) < \epsilon$.

In order to have uniform continuity, we must find a $\delta$ that works when $x \in A$ and $y \in B$.

Suppose that $d(A,B) > 0$, then $\delta = \min\{\delta_1,d(A,B)\}$ works.

Otherwise, $d(A,B) = 0$, and so by compactness $A \cap B$ is non-empty. Then $f$ is continuous on $A \cup B$ (which is compact) by the pasting lemma, and so it is uniformly continuous.

I am not sure whether the conditions can be relaxed further. Perhaps $A$ and $B$ need only be closed?

Answer 3

The function $h$ is not necessarily even continuous.

Let $X=\mathbb{R}, A=\mathbb{Q}, B= \mathbb{R} \setminus \mathbb{Q}$.

Define $f(x) = 1$ for $x \in A$, $g(x) = 0$ for $x \in B$.

Then both $f,g$ are uniformly continuous on $A,B$ respectively, but the function $h$ is not even continuous.

Note: By taking $A=[0,1], B = \mathbb{R} \setminus A$ we obtain the same result, even though $A$ is compact.