Is disjoint union of continuous function continuous too?

Question

So we suppose that $X$ is disjoint union of the collection $$ \mathfrak{X}:=\{X_i:i\in I\} $$ so that I ask if $f_i$ is for any $i\in I$ a continuous function -with respect the subspace toplogy- from $X_i$ to $Y$ then $$ f:=\bigcup \limits _{i\in I}f_i $$ is continuous too. I think that the statement is true when $X_i$ are open because apparently the identity $$ \begin{align}f^{-1}[V] =f^{-1}[V]\cap X =f^{-1}[V]\cap \left (\bigcup \limits _{i\in I}X_i\right )=\bigcup \limits _{i\in I}\left (f^{-1}[V]\cap X_i\right )=\bigcup \limits _{i\in I}\left (f_i^{-1}[V]\cap X_i\right )\end{align} $$ holds, but I am not sure about that. Moreover, what happens if $X_i$ is not always open?

So could someone help me, please?

I am confused. What does it mean for two functions to be "disjoint"? Regarding a function as a collection of ordered pairs, the functions $$f_1 = {(x,y) : y=x} \qquad\text{and}\qquad f_2 = {(x,y): y=x+1}$$ are disjoint, but their union isn't even a function. — Xander Henderson, Jun 25 '22 at 17:18
@XanderHenderson I only want mean that the domain of $f_1$ and $f_2$ are disjoint, that's all. — Antonio Maria Di Mauro, Jun 25 '22 at 17:19
I did not understand---your question is unclear, and I see at least two different ways of interpreting it. Will you please edit your question to fix this ambiguity? Will you also please edit the question to explain some of your own thinking on this question? — Xander Henderson, Jun 25 '22 at 17:22
Now I edited the question: so I know that $f_i:X_i\to Y$ is continuous with respect the subspace topology so that I ask if $\bigcup_{i\in I} f_i$ is continuous when the $X_i$ are disjoint. So did you understand now? — Antonio Maria Di Mauro, Jun 25 '22 at 17:24
No, this is false in general. Let $X=\Bbb{R}$, $X_0=\Bbb{Q}$, $X_1=\Bbb{R}\setminus\Bbb{Q}$ (with standard, subspace, subspace topologies respectively). Let $f_0=0$ and $f_1=1$ be constant functions. Then, these are both continuous with respect to the subspace topologies. However, the 'union' is $f:\Bbb{R}\to\Bbb{R}$, $f(x)=0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational. This function is not continuous (actually nowhere continuous). — peek-a-boo, Jun 25 '22 at 17:36
@peek-a-boo Okay, perfect: however it is true when $X_i$ are open or closed, right? — Antonio Maria Di Mauro, Jun 25 '22 at 17:37
@peek-a-boo I tried to solve it in the question: so is correct my proof attempt? — Antonio Maria Di Mauro, Jun 25 '22 at 17:38
@XanderHenderson: It is not clear to me what you think is unclear about the question. What other interpretation do you see? — Eric Wofsey, Jun 25 '22 at 17:39
the set identities are fine (but I think you have a typo $V,U$ should be the same letter), and the final equality can simply be written as $=\bigcup_{i\in I}f_{i}^{-1}[U]$, since the preimage of $f_i$ is already contained in $X_i$. Now what else can we say about $f_i^{-1}[U]$ if $X_i$ is open in $X$? If $X_i$'s are only assumed closed, there is again an obvious counterexample, closely related to my first comment. (Hint: singletons are closed in $\Bbb{R}$). — peek-a-boo, Jun 25 '22 at 17:43
@peek-a-boo You are right there's a typo: $U$ and $V$ should be the same letter. So if $X_i$ is open then the open set of $X_i$ are the open set of $X$ there contained so that $f^{-1}_i[V]$ is open in $X$ and thus $f^{-1}[V]$ is union of open set so that the proof works for open set (right?) but i do not know if it woks when $X_i$ are closed: so what can you say about? — Antonio Maria Di Mauro, Jun 25 '22 at 17:48
yes that's right in the open case. Regarding the closed case, I suggest you re-read my previous comment, and spend some more time thinking about it (it's not that I don't want to help, but I've already given sufficiently many hints (actually too many if I'm being strict)). — peek-a-boo, Jun 25 '22 at 17:51
@peek-a-boo Oh, do not worry: I will work for a counterexample! Anyway, I think that the statement if true for closed set when $I$ is finite. — Antonio Maria Di Mauro, Jun 25 '22 at 18:05
Oh, eureka!!! Perhaps I found: $\Bbb R$ is disjoint union of its closed singletons and the floor function $\lfloor\quad\rfloor$ is continuous when restriced to ${x}$ for any $x\in\Bbb R$ but it is not continuous, right? — Antonio Maria Di Mauro, Jun 25 '22 at 18:12
correct, the floor function is discontinuous, the function $f$ which I defined above is also discontinuous. Actually, take ANY discontinuous function, and if you restrict to singletons it will be continuous, so you have a counterexample. — peek-a-boo, Jun 25 '22 at 18:36
Your question essentially boils down to "must a piecewise continuous function be continuous". The answer is negative, and there are many, many possible counterexamples. See https://math.stackexchange.com/q/1968943/ . — Xander Henderson, Jun 25 '22 at 19:32
@EricWofsey I am not sure why you feel the need to be so condescending, particularly when I outlined the possible interpretations of the question in my first two comments here. You might also want to look at what the question said when I posted those comments---it has been significantly edited since then. — Xander Henderson, Jun 25 '22 at 19:32
I will also note that there is a further ambiguity---as the question is currently written, I could write that the answer is affirmative. The disjoint union of a family of sets ${X_i}$ is often defined to be the set $$ \bigsqcup_{i \in I} X_i = { (x,i) : i\in I, x\in X_i}. $$ This is a distinct notion from "a union of sets ${X_i}$, where the $X_i$ are disjoint". @AntonioMariaDiMauro Can you please edit your question in order to remove these ambiguities? Leaving clarification in the comments is inadvisable, as comments are ephemeral, and subject to deletion. — Xander Henderson, Jun 25 '22 at 19:42
Assuming that I have understood your question correctly, a correct phrasing might be something like "Suppose that $X_i \subseteq X$ for each $i \in I$, and that the $X_i$ are disjoint. Suppose that for each $i$ there is a continuous function $f_i : X_i \to X$. Define a function $f : \bigcup X_i \to X$ by $$f(x) = \begin{cases} f_i(x) & \text{if $x\in X_i$.} \end{cases} $$ Must $f$ be continuous?" — Xander Henderson, Jun 25 '22 at 19:47
@XanderHenderson: I'm sorry my previous comment came off as condescending--I didn't mean it that way at all, and was genuinely curious what other interpretation you had in mind. Your first comment is about an ambiguity in the meaning of "disjoint functions" but I don't see how that would give rise to an alternative interpretation of the question (only to the title). — Eric Wofsey, Jun 25 '22 at 20:17
@EricWofsey Again, the two functions $${(x,y) : y=x} \qquad\text{and}\qquad {(x,y) : y=x+1}$$ are disjoint sets, each of which satisfies the set-theoretic definition of a function. Under the standard definition of "union", their union is not a function. If one reads the question as literally as it is written, this pretty much answers the question. Indeed, the question is tagged "topology", and these kinds of niggly set-theoretic arguments often show up in intro topology classes. It is not obvious (to me) that "a union of functions" should be understood in any other way. — Xander Henderson, Jun 25 '22 at 20:29
@XanderHenderson: How would that example fit with the question, though? How would $X$ be the disjoint union of the domains of the functions? — Eric Wofsey, Jun 25 '22 at 20:32
@EricWofsey I don't know. Which is why I asked. The question is unclear to me. The fact that one can interpret it in a given way does not mean that this is the interpretation meant by the author. It is the job of the author to be as precise as they can be. — Xander Henderson, Jun 25 '22 at 20:35
@XanderHenderson: But how can one interpret it in that way given that the question directly contradicts that interpretation? As far as I can tell you are just saying the title of the question is ambiguous, but it is very common for titles to have ambiguities that are clarified in the body of the post. — Eric Wofsey, Jun 25 '22 at 20:37
@peek-a-boo Anyway I think that the problem can be solve also using the pasting lemma, right? — Antonio Maria Di Mauro, Jun 26 '22 at 08:32
Every function is a disjoint union of constant functions, which are continuous. — bof, Jun 26 '22 at 08:42
@bof I think your comment is an hint for a counterexample, right? — Antonio Maria Di Mauro, Jun 26 '22 at 08:43

Antonio Maria Di Mauro · Accepted Answer · 2022-06-26T17:46:39.850

The statement is generally false: e.g. the indicator function $\chi _\mathbb{Q}$ of $\mathbb{Q}$ is cotinuous on $\mathbb{Q}$ and $\mathbb{I}$ but it is not continuous on $\mathbb{R}$, which is the disjoint union of $\mathbb{Q}$ and $\mathbb{I}$; moreover, $\chi _\mathbb{Q}$ is continuous in each singleton but it is not cotinuous on $\mathbb{R}$, which is the disjoint union of its singletons, whom are closed, so that the statement is generally false for closed sets too.

However, given $V\in \mathcal{P}(Y)$, we observe that an element $x$ of $\displaystyle \bigcup \limits _{i\in I}\left (f^{-1}[V]\cap X_i\right )$ lies in $X_i$ for any $i\in I$ and is such that $f(x)\in V$ but if $x\in X_i$ then$$f(x)=f_i(x)\in V,$$so that $x$ lies in $\displaystyle \bigcup \limits _{i\in I}\left (f_i^{-1}[V]\cap X_i\right )$; conversely, if $x$ lies in $\displaystyle \bigcup \limits _{i\in I}\left (f_i^{-1}[V]\cap X_i\right )$, then $x$ lies in $f_i^{-1}[V]\cap X_i$ for any $i\in I$, that is, $x$ lies in $X_i$, and in particular$$f(x)=f_i(x)\in V,$$so that $x$ lies in $f_i^{-1}[V]\cap X_i$, that is, in $\displaystyle \bigcup \limits _{i\in I}\left (f^{-1}[V]\cap X_i\right )$. So we conclude that $$ f^{-1}[V] =f^{-1}[V]\cap X=f^{-1}[V]\cap \left (\bigcup \limits _{i\in I}X_i\right ) =\\ \bigcup \limits _{i\in I}\left (f^{-1}[V]\cap X_i\right ) =\bigcup \limits _{i\in I}\left (f_i^{-1}[V]\cap X_i\right ) =\bigcup \limits _{i\in I}f_i^{-1}[V] $$ Now, if $X_i$ is open, then by continuity of $f_i$ surely $f^{-1}_i[V]$ is open in $X$ when $V$ is open in $Y$, so that as the union of open sets is always open, then by the last identity $f^{-1}[V]$ is open in $X$ and so $f$ is continuous; moreover, if $X_i$ is closed, then by continuity of $f_i$ surely $f^{-1}[V]$ is closed in $X$ when $V$ is closed in $Y$ so that as the finite union of closed set is closed, then by the last identity $f^{-1}[V]$ is closed, and so $f$ is continuous.

However, we observe that a similar result holds also when $X_i$ are not disjoint as this relevat result show.

Lemma: Pasting Lemma

If $\varphi$ is a function from a space $X$ to a space $Y$ which is continuous on the sets of an open cover $\mathcal{A}$, then it is globally continuous.

Proof: We observe that for any $V\in\mathcal P(Y)$ the identity $$ \varphi ^{-1}[V]=\varphi ^{-1}[V]\cap X=\varphi ^{-1}[V]\cap \left (\bigcup \limits _{A\in\mathcal{A}}A\right )=\bigcup \limits _{A\in\mathcal{A}}\left (\varphi ^{-1}[V]\cap A\right )=\bigcup \limits _{A\in\mathcal{A}}\left (\varphi \upharpoonright _A\right )^{-1}[V] $$ Now by assumption $\left (\varphi \upharpoonright _A\right )^{-1}[V]$ is open in $A$, but $A$ is open in $X$, hence $\left (\varphi \upharpoonright _A\right )^{-1}[V]$ is open in $X$ and thus as the union of open sets is always open then by the above identity we conclude that $f^{-1}[V]$ is open in $X$, and so $f$ is continuous.

Moreover an analogous result holds if we replace $\mathcal{A}$ with a closed and finite cover $\mathcal{C}$.

So clearly the function $f$ in the question satisfies the hypotheses of the last lemma, so that we conclude that the problem can be solved directely using the Pasting Lemma.

Is disjoint union of continuous function continuous too?

1 Answers1