# The continuity of a function defined on disjoint union.

Let $$X\subset \mathbb R$$ and $$f:X\to \mathbb R$$ be a function.

I want to show

If $$A,B\subset \mathbb R$$ satisfy $$X=A\cup B, A\cap B=\emptyset,$$ then

$$f$$ is continuous $$\iff$$ $$f|_A$$, $$f|_B$$ are both continuous.

I did $$\Rightarrow$$. So I have to show opposite.

Suppose $$f|_A$$, $$f|_B$$ are both continuous.

To show the continuity of $$f:X\to \mathbb R$$, I'll show $$\forall O\in \mathcal O_{\mathbb R}\ ; \ f^{-1}(O)\in \mathcal O_{\mathbb R}^X,$$ where $$\mathcal O_{\mathbb R}$$ is Euclidean topology and $$\mathcal O_{\mathbb R}^X$$ is relative topology, i.e., $$\mathcal O_{\mathbb R}^X=\{ X\cap O \mid O\in \mathcal O_{\mathbb R} \}.$$

I have $$f^{-1}(O)=(f|_A)^{-1}(O)\cup (f|_B)^{-1}(O).$$

And from the continuity of $$f|_A$$ and $$f|_B$$, I have $$(f|_A)^{-1}(O)\in \mathcal O_{\mathbb R}^A,\ (f|_B)^{-1}(O)\in \mathcal O_{\mathbb R}^B.$$ So there exists $$O_1, O_2\in \mathcal O_{\mathbb R}$$ s.t. $$(f|_A)^{-1}(O)=A\cap O_1,\ (f|_B)^{-1}(O)=B\cap O_2.$$

Then, \begin{align} f^{-1}(O) &=(f|_A)^{-1}(O)\cup (f|_B)^{-1}(O)\\ &=(A\cap O_1) \cup (B\cap O_2) \end{align}

So, to show $$f^{-1}(O)\in \mathcal O_{\mathbb R}^X$$, I want to say $$(A\cap O_1) \cup (B\cap O_2) =X\cap [\mathrm{an \ element \ of\ } \mathcal O_{\mathbb R}]. \ \cdots (\ast)$$

But I'm stacked here. I don't know how I should show $$(\ast)$$.

• That is false, see for instance $X=\Bbb R$, $A=(-\infty,0)$, $B=[0,\infty)$ and $f(x)=\begin{cases}0&\text{if }x\ge 0\\ 1&\text{if }x<0\end{cases}$. Oct 27, 2022 at 16:44
• In topology "disjoint union" does not mean the union of disjoint subspaces, but the union of disjoint $A, B$ which are both open in $A \cup B$. Oct 27, 2022 at 16:50
• The question clearly indicates that $A, B$ are disjoint sets. And then it's false, as Sassatelli Giulio's example shows.
– Ulli
Oct 27, 2022 at 17:48
• Of course, if $A, B$ are disjoint, open sets its true. But there is a more powerful version: It suffices that $A, B$ are closed sets, whose union is $X$ (not necessarily disjoint). Try to prove this one, it is really helpful in many situations.
– Ulli
Oct 27, 2022 at 18:19

$$\Longleftarrow$$ is false. In his comment Sassatelli Giulio gave the counterexample $$X = \mathbb R, A = (-\infty,0), B= [0,\infty), f(x) = \begin{cases} 1 & x \in A \\ 0 & x \in B\end{cases}$$.

But, as Paul Frost comments, perhaps you misunderstood the concept of disjoint union in topology. A space $$X$$ is the disjoint union of subspaces $$A, B$$ if

1. $$A \cup B = X$$.
2. $$A \cap B = \emptyset$$.
3. $$A, B$$ are open in $$X$$.

Note that 3. is equivalent to the requirement that $$A, B$$ are closed in $$X$$.

In your question you consider the case $$X \subset \mathbb R$$. Condition 3. does not mean that $$A, B$$ are open (or closed) in $$\mathbb R$$. An example is $$X = (0,1] \cup [2,3), A = (0,1], B = [2,3)$$.

Now forget that $$X \subset \mathbb R$$. We only need to know that it is a topological space.

To show that $$f$$ is continuous, we have to show that for $$f^{-1}(U)$$ is open in $$X$$ for each open $$U \subset \mathbb R$$. We know that $$(f \mid_A)^{-1}(U) = f^{-1}(U) \cap A$$ is open in $$A$$ and $$(f \mid_B)^{-1}(U) = f^{-1}(U) \cap B$$ is open in $$B$$. But $$A$$ and $$B$$ are open in $$X$$, thus also $$f^{-1}(U) \cap A$$ and $$f^{-1}(U) \cap B$$ are open in $$X$$. Therefore $$f^{-1}(U) = f^{-1}(U) \cap A \cup f^{-1}(U) \cap B$$ is open in $$X$$.