Find a surjective closed mapping that not an open mapping and a surjective open mapping that is not a closed mapping

Question

Find a surjective closed mapping that not an open mapping and a surjective open mapping that is not a closed mapping

Attempt:

Suppose $X$ and $Y$ are metric spaces and $f : X \rightarrow Y$. We call $f$ an open mapping if and only if for each open subset $U$ of $X$,the image $f(U)$ is open in $Y$ and we call $f$ a closed mapping if and only if for each closed subset $F$ of $X$, the image $f(F)$ is closed in $Y$.

How do I move next. Searching for an example seems a bit like trial and error to me. Could there be a logical way of deducing such examples?

Answer 1

Closed, not open. Define

$$f(0)= 0,\ f(1/4) = 1/2, \ f(1/2) = 0, \ f(1) = 1.$$

(Good to draw a picture.) So we have four points on the graph of $f.$ Connect these points in order with line segments. We then have a piecewise-linear, continuous and surjective $f: [0,1]\to [0,1].$ If $E\subset [0,1]$ is closed, then $E$ is compact, hence $f(E)$ is compact by continuity and therefore closed. But notice $f((0,1/2)) = (0/1/2],$ hence $f$ is not an open map.

Open, not closed: This is kind of a wild example. Suppose $I_1, I_2, \dots$ are the open intervals with rational endpoints. Then we can inductively choose pairwise disjoint Cantor sets $K_n\subset I_n.$ Recall each $K_n$ has the cardinality of $\mathbb {R}.$ On $K_1$ we define $f_1$ so that $f_1(K_1) = (0,1).$ For $n>1,$ we choose any $f_n$ on $K_n$ such that $f_n(K_n) = \mathbb {R}.$ Now define $f$ on all of $\mathbb {R}$ by setting $f = f_n$ on $K_n, n = 1,2,\dots,$ and defining $f=0$ everywhere else. We arrive at a surjective $f:\mathbb {R} \to \mathbb {R}.$ This $f$ is an open map: If $U$ is open in $\mathbb {R},$ then $U$ will contain some $I_n$ for $n>1,$ hence $f(U)$ contains $f(K_n) = \mathbb {R}.$ Therefore $f(U) = \mathbb {R},$ which of course is open in $\mathbb {R}.$ But $f$ is not closed, because $K_1$ is closed and $f(K_1) = (0,1),$ which is not closed in $\mathbb {R}.$