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.

1 (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. )

  • $\begingroup$ That function is continuous. You don't take the preimage by computing the inverse function at each point. As you noted, that function has no inverse, but that's not a problem. The inverse image of $(-1,1)$ is a disjoint union of two open intervals hence is open. $\endgroup$ – Matt Samuel Jul 18 '16 at 21:08
  • $\begingroup$ I see, i have a another candidate I'm working on. I'll edit the question soon. Thanks. $\endgroup$ – Michael Maliszesky Jul 20 '16 at 16:44
  • $\begingroup$ For the current version: you're right that the function is both open and not continuous; however, it's also closed, which isn't what you want. Indeed, this function is a (set) bijection, and its inverse is continuous (you can check), which means the function itself is both open and closed. $\endgroup$ – Greg Martin Jul 20 '16 at 22:27
  • $\begingroup$ @Greg Martin, I just added a reason why I think to function is not closed. It based on the same reasoning that shows $f(x) = \frac{1}{1+x^2}$ is not closed. Is there something wrong with my reason? $\endgroup$ – Michael Maliszesky Jul 21 '16 at 20:51
  • $\begingroup$ Yes, something subtle. Forget the second coordinate for a second, and consider these two maps: $f\colon[0,\infty)\to(0,1]$ defined by $f(x)=\frac1{1+x^2}$, and $g\colon[0,\infty)\to[0,1]$ defined by $g(x)=\frac1{1+x^2}$ (all spaces with the subspace topology from $\Bbb R$). Then $f$ is a closed map, but $g$ isn't a closed map. The difference is that the image $(0,1]$ is closed as a subset of $(0,1]$, but isn't closed as a subset of $[0,1]$. $\endgroup$ – Greg Martin Jul 21 '16 at 22:56

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 $X.$

  • $\begingroup$ I have one question, why can we use the definition of continuity in $\mathbb{R}$ to show that it is discontinuous at 0? If we would restrict ourselves to working in X and using the "preimage of an open set is open" definition of continuity in X, I fail to see how we could create an open set U such that $f^{-1}(U)$ is not open in X. If we would take your example and look at a set U such that $f^{-1}(U) = \{0\}$, then $U = \{1\} \cup \{0\}$ which is open in X. So my question is, I guess, if we look at the elements $(1/n)$ such that $f(1/n) \rightarrow 0$, why are they not contained in X? $\endgroup$ – eager2learn Oct 6 '18 at 12:40
  • $\begingroup$ @eager2learn The two definitions are the same. Note $\{1\}$ is open in $X$ but $f^{-1}(\{1\})=\{0,1\}$ is not open in $X.$ $\endgroup$ – zhw. Oct 6 '18 at 16:13

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