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Let $X=A\cup B$ and $A,B$ closed. Let $f: A\to Y$ and $g: B\to Y$ be continuous functions and $f(x)=g(x)$ for all $x\in A\cap B$. Show that $h: X\to Y$, $h(x)=\begin{cases} f(x), \text{if}\,\,x\in A \\ g(x), \text{if}\,\, x\in B\end{cases}$

is well-defined and continuous.

To show, that $h$ is a well-defined function, should be easy. We have to show, that $h$ is 'left-totally'. Hence for every $x\in X$ exists a $y\in Y$ with $h(x)=y$.

And 'right-unique'. Hence for every $x\in X$ and $y, y'\in Y$ with $h(x)=y$ and $h(x)=y'$ is $y=y'$.

It is immediatly clear, that $h$ is 'left-totally' since $X=A\cup B$ and the definition of $h(x)$.

Also it is immedialty clear, that $h$ is 'right-unique', because $f$ and $g$ are functions (which are 'right-unique'). You could seperate some cases. For example: If $x\in A\setminus B$ and $h(x)=y$ and $h(x)=y'$. Then is $h(x)=f(x)=y$ and $f(x)=y'$. But $f$ is 'right-unique' and therefor $y=y'$.

The same for $x\in B\setminus A$ and $x\in A\cap B$. Actually, I do not even know, why this is a task. Maybe just to refresh what is meant with 'well-defined', or is there something to think about here?

And $h$ is continuous, because it is continuous in every point.

Am I missing something here? What has this to do with $A, B$ beeing closed?

Thanks in advance.

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Cornman
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  • Well, why is it continuous at every point? Closedness is actually needed for this to be true, so you've got more work to do. The part about being a well defined function is indeed easy, though a complete argument needs of course to use the fact that $f$ and $g$ agree on the intersection. – Kevin Carlson Feb 25 '19 at 01:00
  • What do "left-totally" and "right-unique" mean? Never seen these words before. And how did you prove continuity? – stressed out Feb 25 '19 at 01:01
  • @stressed-out the term should be left-total, but otherwise they were both defined in the text. – Kevin Carlson Feb 25 '19 at 01:02
  • @KevinCarlson Yes, thanks. You're right that both of them have been defined in the text. And I can see why it's the same as the standard definition of being well-defined. – stressed out Feb 25 '19 at 01:03
  • @stressedout: What would be the actual terms? – Cornman Feb 25 '19 at 01:05
  • @KevinCarlson Is it easier to show, that $h$ is continuous at every point, or that the preimage of open/closed subsets is open/closed? – Cornman Feb 25 '19 at 01:07
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    I think $A$ and $B$ are assumed to be closed because of this: since $f: A \to Y$ and $g: B \to Y$ are continuous, the function $H: A \cap B \to Y$ defined by $H(x) = f(x) - g(x)$ is continuous. Since the set $Z={x \in A \cap B: f(x) = g(x)}$ is $H^{-1}({0})$, it is closed in $A \cap B$. Since $A \cap B$ is closed in $X$, $Z$ is also closed in $X$. But this, of course, does not prove that the assumption of $A$ and $B$ being closed is necessary. It's just a mere observation which can also give some ideas about the proof, possibly. – stressed out Feb 25 '19 at 01:12
  • Also, I was unfamiliar with your terminology. They're fine. Except that according to Kevin Carlson, you had to write "left-total" instead of "left-totally" which makes sense. – stressed out Feb 25 '19 at 01:14
  • It turns out that the assumption that $A,B$ are closed in $X$ comes to our assistance when we want to show that something closed in $A$ or $B$ must be closed in $X$ too. – stressed out Feb 25 '19 at 01:45
  • @Cornman I would say the latter. – Kevin Carlson Feb 25 '19 at 02:26

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Showing that $h$ is well-defined is straightforward and you've done it already. Here's the part about continuity:

A well-known theorem in general topology, which is straightforward to prove, says that

A function $\varphi: X \to Y$ between two topological spaces $X$ and $Y$ is continuous if and only if the pre-image of any closed set in $Y$ is closed in $X$.

Now let $Y_0 \subseteq Y$ be a closed set in the topological space $Y$. Then what is $h^{-1}[Y_o]$? Since $X= A \cup B$, one can write set-theoretically that

$$h^{-1}[Y_0]=\big\{x\in A: f(x)\in Y_0\big\}\cup\big\{x\in B: g(x)\in Y_0\big\}$$

which tells us that $$h^{-1}[Y_0] = f^{-1}[Y_0]\cup g^{-1}[Y_0]$$

Since $f$ is continuous, $f^{-1}[Y_0]$ is closed in $A$, since $A$ itself is closed in $X$ (this is where the assumption that $A$ and $B$ are closed in $X$ becomes useful), $f^{-1}[Y_0]$ is closed in $X$.

A similar argument for $B$ shows that $g^{-1}[Y_0]$ is closed in $X$. Hence, $h^{-1}[Y_0]$ is closed in $X$. Q.E.D.

A possible reason that this exercise was given is to emphasize that being open or closed are relative concepts. To move from being closed in a subspace to a superset is not trivial and one needs extra assumptions to make this move. Here, the assumption that $A$ and $B$ are closed in $X$ allows us to conclude that $f^{-1}[Y_0]$ and $g^{-1}[Y_0]$, not only are closed in $A$ and $B$ respectively, but they're closed in $X$ as well.

Addendum: In fact, we have the following characterization of closed sets in the subspace topology.

Proposition. Let $Y$ be a subspace of a space $X$. Then a set $H\subseteq Y$ is closed in $Y$ if and only if $H=F\cap Y$ for some closed set $F$ in $X$.

To see the proof, which follows the same line of thought as in your comment, see Brian M. Scott's post on this question.

stressed out
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  • One more question: $f^{-1}[Y_0]$ is closed in $A$, since $A$ is closed in $X$, $f^{-1}[Y_0]$ is closed in $X$. – Cornman Feb 25 '19 at 04:13
  • Why is that? Is it because of the subspace topology? I write $f^{-1}[Y_0]=Y_0'$. Then $(Y_0')^c$ (the complement) is open in $A$. $A$ is a topological space with regards to the subspace topology. This means there is an open set $U\subseteq X$ with $U\cap A=(Y_0')^c$. And this $U$ equals(?) $(Y_0')^c$. Which means, that $(Y_0')^c$ is open in $X$ and therefor $Y_0'=f^{-1}[Y_0]$ is closed in $X$? I wonder because at this stage in the book, the subspace topology is not defined yet. – Cornman Feb 25 '19 at 04:18
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    @Cornman Yes, you're right that it is because of the subspace topology. I'm afraid there's no way to bypass subspace topology because the definition of something being closed in a subset of the original space is pretty much possible only because of the subspace topology. Also, your argument looks OK to me except the part that you conclude $U=(Y'_0)^c$. I haven't checked that part, but your idea seems to be on the right track but needs to be polished. See the link I added in my answer for a well-written "if and only if" proof by Brian M. Scott. – stressed out Feb 25 '19 at 12:39
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    Thanks, that is exactly what I thought too. – Cornman Feb 25 '19 at 14:44