Why the following function is closed?

Let $$X=[0,1] \cup [2,3]$$ and let $$Y=[0,2]$$. I have $$p: X \to Y$$ defined by

$$p(x)= \begin{cases} x \quad &\text{for} \ x \in [0,1] \\ x-1 \quad &\text{for} \ x \in [2,3] \end{cases}$$

I have to prove that $$p$$ is a closed map. I tried a lot of times, but I didn't get it. And I have to prove it without compactness

• Choose a closed subset of $Y$ and calculate the inverse image. Show that this is always closed. Make sure to take care around $p^{-1}(1)$. Sep 7 '20 at 22:03
• @CyclotomicField How does that prove $p$ is a closed map? You need to show that for all closed $C\subseteq X$, $p(C)\subseteq Y$ is closed. Sep 7 '20 at 22:07
• @JustinYoung OP wants to do prove it without compactness Sep 7 '20 at 22:18

Hint: you have to show that if $$A \subseteq [0, 1] \cup [2, 3]$$ is closed, then $$p(A)$$ is closed. Now $$p(A) = p(A \cap [0, 1]) \cup p(A \cap [2, 3])$$, which will be closed if $$p(A \cap [0, 1])$$ and $$p(A \cap [2, 3])$$ are both closed (that's sufficient, but not necessary, but it will do for this problem). $$A$$ is closed iff $$A \cap [0, 1]$$ and $$A \cap [2, 3]$$ are both closed (do you see why?). Now $$p$$ restricts to the identity on $$[0, 1]$$ and to the function $$x \mapsto x - 1$$ on $$[2, 3]$$. I'll let you take it from there, by showing that both of these restrictions of $$p$$ are closed functions.
Hint: Let $$A\subseteq X$$ be a closed set. Show that $$p(A)$$ is a closed set. If $$\{x_n\}_n\subseteq A$$ converges to some $$x\in A$$ then we have $$\{p(x_n)\}_n\subseteq p(A)$$. $$p$$ is continuous (on the appropriate tail of the sequence) thus we must have $$\lim p(\{x_n\})=p(\lim\{x_n\})=p(x)$$. But since $$x\in A$$ then $$p(x)\in p(A)$$.
• I'm really cutting corners with the above argument. One would also need to take care of the case where your closed set $A$ is a union of say two closed sets, one in $[0,1]$ and one in $[2,3]$. Sep 7 '20 at 22:12