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I came across this exercise, the main problem for me are the restrictions, i need to find examples for maps $f: \mathbb {R}^2 \rightarrow \mathbb {R}^2$ or subsets of them such that $f$ is only one or two of the three at the same time. (I.e $f$ is open but no closed nor continuous) But i don't know How to build such examples, as i also don't see the kind of patologies i should be looking for

Daniel Moraes
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  • Do you know any of them? – SmileyCraft Jan 21 '19 at 22:30
  • On open mappings that are not continuous: "Explicit examples are moderately difficult to construct, but it’s not too hard to come up with non-constructive examples; here’s one such [...]" https://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous – Nap D. Lover Jan 21 '19 at 22:38
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    @MoisheCohen There is an important difference as that question asks for $f:X\to Y$ where $X,Y\subseteq\mathbb{R}^2$ while this question requires $f:\mathbb{R}^2\to\mathbb{R}^2$. – SmileyCraft Jan 21 '19 at 22:55
  • @SmileyCraft: True, but two out of three examples given there are defined on $R^2$ and the remaining example of continuous but not closed is only superficially defined on a smaller subset since the open disk is homeomorphic to $R^2$. – Moishe Kohan Jan 21 '19 at 23:11
  • @MoisheCohen Well I have been looking around on the internet and I honestly can not find any open closed function $f:\mathbb{R}^2\to\mathbb{R}^2$ that is not continuous. I have found an open function $f:\mathbb{R}^2\to\mathbb{R}^2$ that is not continuous, or rather LoveTooNap29 did, but it is not closed. All the other 6 combinations of open, closed and continuous are rather simple. – SmileyCraft Jan 21 '19 at 23:17
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    @SmileyCraft i actually tried to say that subsets would work – Daniel Moraes Jan 21 '19 at 23:31

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This is all I currently know about all eight cases.

Open, closed, continuous: $f(x,y)=(x,y)$

Open, closed, not continuous: Very interesting A function $f:\mathbb{R}^2\to\mathbb{R}^2$ that is open and closed, but not continuous.

Open, not closed, continuous: $f(x,y)=(e^x,y)$

Open, not closed, not continuous: Any strongly Darboux function.

Not open, closed, continuous: $f(x,y)=(0,0)$

Not open, closed, not continuous: $f(x,y)=(\lfloor x\rfloor,0)$

Not open, not closed, continuous: $f(x,y)=(x,0)$

Not open, not closed, not continuous: $f(x,y)=(x+\lfloor x\rfloor,0)$

If you simply want an open, closed function $f:A\to B$ that is not continuous, where $A,B\subseteq\mathbb{R}^2$, then $f(x,y)=(x-\lfloor x\rfloor,y):\mathbb{R}^2\to\{(x,y)\in\mathbb{R}^2:x\in[0,1)\}$ suffices.

SmileyCraft
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    I don't get the open not closed not continuous case, i was trying for product topology but $sin(x)$ does not take open intervals to open intervals necesseraly, i,e. $sin(A)=[0,2π]$ for some $A$ – Daniel Moraes Jan 22 '19 at 11:24
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    Good point. I was already hesitant on my finding. Unfortunately, the simplest example is then probably any strongly Darboux function. – SmileyCraft Jan 22 '19 at 11:42
  • Don't know It but Will look It up – Daniel Moraes Jan 22 '19 at 11:52
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A constant map is continuous, closed and not open.

Map the closed upper half of the plane to (1,1) and the open lower half to (0,0).
That map is not continuous, closed, not open.

William Elliot
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If $f: A\to B$ is a continuous surjection, and $B$ is a subspace of $C,$ and $C$ is a subspace of $D,$ such that $B$ is not closed in $C$ and $B$ is not open in $D,$ then $f:A\to D$ is continuous and $f$ maps $A,$ which is an open-and-closed subset of $A$(!) onto $B,$ which is neither open nor closed in $D.$

For example $A=\Bbb R^2,\,B= (-\pi/2, \pi/2)\times \{0\},\, C=\Bbb R\times \{0\},$ and $D=\Bbb R^2.$ With $f(x,y)=<\arctan x,\,0>.$

( I am using round brackets $(...,...)$ in the definition of $B$ to denote an open real interval, and $<...,...>$ in the definition of $f$ to denote an ordered pair.)

Part of the idea is to start with $A,B,$ and $f,$ and look for $C$ such that $B$ is a dense proper subset of $C,$ and $C$ has empty interior in $D$. This is made easier by taking $1$-dimensional $B$ and $C$, and $D=\Bbb R^2$.

DanielWainfleet
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