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

Question

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.

Answer 1

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.

Answer 2

Here is one solution to this exercise, which relies only on a few key insights. Below, $X$ and $Y$ are topological spaces, and $S \subseteq X$.

First, consider the inclusion map $i: S \to X$, which is always continuous by definition of the subspace topology. By carefully choosing $S$, we can control whether $i$ is open or closed:

• If $S$ is open but not closed, then $i$ is open but not closed.
• If $S$ is closed but not open, then $i$ is closed but not open.
• If $S$ is neither closed nor open, then $i$ is neither closed nor open.

Next, consider the map $g : \mathbb{S}^1 \to [0, 1), g(e^{2\pi i\theta}) = \theta$. This map is the inverse of the exponential map $f : [0, 1) \to \mathbb{S}^1, f(\theta) = e^{2\pi i\theta}$, which is a continuous bijection from $[0, 1) \to \mathbb{S}^1$ but is not a homeomorphism. Thus,

• $g$ is open and closed (because images under $g$ correspond to preimages under $f$) but not continuous (else $f$ would be a homeomorphism).

Now, we may compose $g$ with carefully chosen inclusion maps to cover the remaining cases:

• $g : \mathbb{S}^1 \to [0, \infty)$ open but neither continuous nor closed (because the inclusion $[0, 1) \to [0, \infty)$ is open but not closed).
• $g : \mathbb{S}^1 \to (-\infty, 1)$ closed but neither continuous nor open (because the inclusion $[0, 1) \to (-\infty, 1)$ is closed but not open).