An example of open but not closed map.

Question

Exercise: Find an example of mapping which is open but not closed, which is closed but not open.

I am thinking of trivial examples with $X=\{1,2,3\}$ however I have no idea on how to build a function that preserves the open interval but no the closed ones. Since this is my first exercise of this kind.

Question:

Can someone give me a hint?

Thanks in advance!

A function which is both open and bijective must also be closed - so any example must be a function which is not bijective. (The existing answers do include examples where the function is injective but not surjective, as well as examples where the function is surjective but not injective.) — Daniel Schepler, Dec 05 '18 at 18:23

cqfd · Answer 1 · 2018-12-05T18:13:17.880

15

Consider the projection mapping $f:\Bbb R^2\to \Bbb R$ defined as $f(x_1,x_2)=x_1$. $f $ is continuous and open, but not closed. (Consider the image of the hyperbola $x_1x_2=1$ under $f $.)

edited Dec 05 '18 at 18:13

answered Dec 05 '18 at 17:40

cqfd

12,974

1

can you please prove rigorously why this map is not closed? I am not able to get it!! – Jun 16 '20 at 18:20
2

@user795826 Draw a graph of the hyperbola and project onto the $x_1$-axis. What do you get? – cqfd Jun 17 '20 at 03:20
I understand what I will get by projecting hyperbola on $x_{1}$ but I don't know how it proves that it's not closed map. Can you please elaborate? – Jun 17 '20 at 03:38
3

@user795826 The image is just $\mathbb R\setminus{0}$. It is not closed since $(1/n)_{n\geq1}\subseteq \mathbb R\setminus{0}$ is a converging sequence in $\mathbb R$ that doesn't have a limit in $\mathbb R\setminus{0}$. – cqfd Jun 17 '20 at 05:10

score 4 · Accepted Answer · answered Dec 05 '18 at 17:36

4

Starting with the simplest example of maps from $\mathbb R$ into itself, which are the constant maps, would be a good idea.

On the other hand, if $f$ is a map from $\mathbb R$ into itself, and if its image is a subset of $\mathbb R$ which is not closed, then $f$ cannot possibly be a closed map.

answered Dec 05 '18 at 17:36

José Carlos Santos

440,053

I was thinking about the constant maps. Would $f(x)=5$ be a constant map? – Pedro Gomes Dec 05 '18 at 18:09
Yes, of course. Is it closed? Is it open? – José Carlos Santos Dec 05 '18 at 18:10
I would guess it is closed since ${5}$ is a singleton in $\mathbb{R}$ with the standard topology. – Pedro Gomes Dec 05 '18 at 18:12
Yes, you are right. Add to that that, in $\mathbb R$ with the standard topology, (or in any metric space) every singleton is closed. – José Carlos Santos Dec 05 '18 at 18:14
@JoséCarlosSantos can you please tell example of a map which is open but not closed? I think that was not covered in your answer. – Jun 16 '20 at 16:19
@User Take any injective map from $\Bbb R$ into $\Bbb R$ which is not surjective (such as $\arctan$, for instance). – José Carlos Santos Jun 16 '20 at 16:22
@JoséCarlosSantos thanks a lot!! – Jun 16 '20 at 17:57

score 3 · Answer 3 · answered Dec 05 '18 at 17:35

3

Hint: Consider the inclusion functions $U \hookrightarrow \mathbb{R}$ for suitable subsets $U$ of $\mathbb{R}$, giving $\mathbb{R}$ its usual Euclidean topology, but giving $U$ the indiscrete (trivial) topology.

What happens if $U$ is open but not closed in $\mathbb{R}$? What happens if $U$ is closed but not open in $\mathbb{R}$?

answered Dec 05 '18 at 17:35

Clive Newstead

66,200

Why cannot a set inside $U$ be mapped to a set that is not necessarily open(considering the first case)? – Pedro Gomes Dec 05 '18 at 20:41
@PedroGomes: This is why I said to equip $U$ with the indiscrete topology, because then the only open subsets of $U$ are $\varnothing$ and $U$ (and the only closed subsets are $\varnothing$ and $U$). Of course, $U$ has other subsets which are open or closed in the Euclidean topology, but that doesn't matter. – Clive Newstead Dec 05 '18 at 20:49
Could you give a concrete example please? Like for example an interval? – Pedro Gomes Dec 05 '18 at 22:21
@PedroGomes: $U = (0,1)$ is open in $\mathbb{R}$ but not closed in $\mathbb{R}$; and $U = [0,1]$ is closed in $\mathbb{R}$ but not open in $\mathbb{R}$. – Clive Newstead Dec 06 '18 at 13:29

score 3 · Answer 4 · answered Aug 07 '23 at 08:10

Open not closed; $X=\{a,b,c\}$ , $\tau=\{\emptyset, X,\{a\}\}$ then take $f:(X,\tau)\to (X,\tau)$, $f(x)=a $ the constant function which maps all open sets to open set {a}. But for the closed set $\{b,c\}\mapsto \{a\}$ and $\{a\}$ is not closed.

Closed not open; in $\mathbb{R}$ with usual topology take $f(x)=a$ constant function, which maps all open sets to a singleton and singleton sets are closed in usual topology of $\mathbb{R}$. And maps all closed sets to closed set {a}.

It is the same constant function in both examples. Note that the domain, codomain and topology is important not the definition of a function.

score 1 · Answer 5 · answered Dec 05 '18 at 17:54

Any constant function $f:\mathbb{R} \to \mathbb{R}$ is closed but not open. For an open but not closed map, let $g:\mathbb{R} \to\mathbb{R}$ denote the Conway Base 13 function—a pathological function which maps all nonempty open sets to $\mathbb{R}$—and compose it with any map $f:\mathbb{R} \to \mathbb{R}$ whose image is precisely $(0,1)$. The resultant function is open because all nonempty open sets are mapped to $(0,1)$, but not closed because, for instance, the image of $[0,1]$ is $(0,1)$ which is not closed in $\mathbb{R}$.

An example of open but not closed map.

5 Answers5