I'm trying to teach my self topology. The book I'm using has the following problem:
Give an example of two subsets $X,Y \subseteq \mathbb R ^2$, both considered as topological spaces with their Euclidean topologies, together with a map $f : X \rightarrow Y$ that has the property:
* $f$ is open but neither closed not continuous.
I basically have 2 questions:
(1)I have an example that I'm pretty sure is good. Is it a good example and how can I improve my justification?
(2) can anyone please come up with another example of a surface to surface map?
(my example) $X = [0, \infty) \times [-2,2] $ and $Y = [0,1] \times [-3,1) \cup \{0\} \cup (1,3] $, also $f(x,y) =( \frac{1}{x^2+1}, y+\frac {y}{|y|}).$ *Or y=0
*When $y=0$: $f(x,0)=(\frac{1}{x^2+1},0)$.
(not closed) Let $V = [0,\infty) \times [\frac{1}{2},\frac{3}{2}]$. Then $f(V)=(0,1]\times[\frac{3}{2},\frac{5}{2}]$. V is closed, f(V) is not closed. (Shown as the green/blue faded rectangle below)
(not continuous) Let $U = B_{\frac{1}{4}}(\frac{1}{2},0)$ (the open ball centered at (0.5,0) with r = 0.25). This is considered open in $Y$'s space. $f^{-1}(U)=(\frac{1}{\sqrt{3}}, \sqrt{3} )\times \{0\}$ is a line segment on the x-axis, and this is not open in $X$'s space. (Shown below)
(is open) I'm not sure how to prove its open, but I'm pretty sure it is.
(The image above was made with a program I wrote, the actual domain used in the program goes from 0 to 9 on the x-axis and -2 to 2 on the y-axis. The actual function goes to infinity on the x-axis, but all x values greater than 4 get compressed to about the size of 1 pixel, so there is no need to draw beyond that. The y-axis is labeled between -3 to 3, but only values between -2 and 2 are mapped. I made to original surface pink, so white space is empty. )
Answer to (2): Let $X=Y= \{1/n: n = 1,2,\dots \} \cup \{0\}.$ (These are subsets of $\mathbb R,$ but we can view them as subsets of $\mathbb R^2.$) Define $f:X\to X$ by setting $f(0)= 1, f(1/n) = 1/n, n = 1,2,\dots.$ Claims: $f$ is open, $f$ is discontinuous at $0,$ and $f$ is not closed. The first claim follows by understanding what the open subsets of $X$ look like, the second claim is clear, and the third claim is seen by noting $f(X) = X \setminus \{0\},$ which is not closed in $X.$