# Maps from subsets of $\mathbb{R}^2$ that are either open/closed/continuous

I'm self studying J. Lee's Introduction to Topological Manifolds, and after doing all other exercises on chapter 2, I can't seem to come with the proper counterexamples for this one.

For each of the following properties, give an example consisting of two subsets $X, Y \subset \mathbb{R}^2$, both considered as topological spaces with their Euclidean topologies, together with a map $f: X \rightarrow Y$ that has the indicated property.

• Open but not closed/continuous

• Closed but not open/continuous

• Continuous but not open/closed

etc... (all cases)

I think the reason for my difficulty is a lack of geometric intuition of what open, closed and continuous maps look like as deformations of shapes in $\mathbb{R}^2$ (I can construct counterexamples from artificial made up spaces).

I tried squashing balls and 'pumping space' into lines, as well as joining strips in annulus shapes, but all the examples I seem to come up with are either too well behaved (i.e. all 3) or too misbehaved.

Instead of a case by case answered I would prefer a couple worked out cases and some intuition on how to visualise open / closed maps in this context.

Here I give you some easy to visualize examples.

Consider $X=\{(x,y)\in\mathbb{R}^2/x^2+y^2<1,\}$ which is an open ball of radius $1$ and $Y=\{(x,y)\in\mathbb{R}^2/x^2+y^2<4,\}$ which is an open ball of radius $2.$ The inclusion map is open but not closed, because the image of $X$ is not closed in $Y.$

Consider any $X$ and $Y$ and a constant map. It is closed, because a set consisting of a single point is closed, but not open. It is also continuous but not open.

A continuous but not closed map is $f:\mathbb{R}^2\rightarrow \mathbb{R}^2, f(x,y)=(\frac{1}{1+x^2+y^2},0).$ The image of the closed set $\mathbb{R}^2$ is $(0,1]\times{0}$ which is not closed.

The function $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ given by $f(x,y)=\left\{\begin{array}{ccc}(0,0) & \mbox{if} & x<0 \\ (1,1) & \mbox{if} & x\geq 0 \end{array}\right.$ is closed but not continuous. It is closed because the image of any (closed) set is $\{(0,0)\},$ $\{(1,1)\},$ or $\{(0,0),(1,1)\},$ which are closed sets.

• For completeness, you may want to add example of a map $X\to Y$ which is open but not continuous (same for closed but not continuous). There is even a map of this type ${\mathbb R}\to {\mathbb R}$, but this is harder. – Moishe Kohan May 26 '14 at 1:14
• @studiosus What is the example of a closed map that's not continuous? Especially ${\mathbb R}\to {\mathbb R}$. – user4894 May 26 '14 at 6:02
• Map whose image is finite but more than one point. – Moishe Kohan May 26 '14 at 15:32