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Is this question acceptable?: Can the graph of a continuous function be rotated to become the graph of a discontinuous function?

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    http://mathoverflow.net/questions/49027/can-a-graph-be-both-continuous-and-discontinuous

    There seems to be some confusion about the meaning of the OP. I take the question to mean

    Can there exist two functions $f,g: \mathbb R \to \mathbb R$ so that $f$ is continuous, $g$ is discontinuous, and their graphs $\Gamma_f, \Gamma_g \subseteq \mathbb R^2$ are related by a rotation?

    Under this interpretation, I don't see why the question should be closed. I propose to edit the question to have to above statement and reopen. What do you think?

    Supposing I edit the question, I think the [gn.general-topology] tag is appropriate, but [ca.classical-analysis] seems wrong. Are there other appropriate tags?

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    This is also the way I understand the question, and I don’t see why it should be closed.
    And perhaps replace “rotation” by “isometry” in the corrected question.

    About the tags, I don’t see other tag than [gn.general-topology].
    • CommentAuthorWillieWong
    • CommentTimeDec 11th 2010
     

    @Anton: I voted to close the first version of the question. Even in the second incarnation I agree more or less with fedja. I have two problems with your proposition:

    (a) I prefer to leave the clarification up to the OP, rather than putting words in his mouth.

    (b) I don't see how the question is interesting even in your current re-write. Perhaps I'm being slow, but if $S$ is the graph of a continuous function $f$, then $S$ is a connected set. And isn't it the case that if $S'$ is the graph of a discontinuous function, then $S'$ is a disconnected set. If so then as stated it is somehow trivially false. Perhaps a better re-writing would be

    Does there exist $f$ and $g$ .... such that the closure of their epigraphs $U_f , U_g$ are related be an isometry?

    If someone can explain to me why there is a subtlety in the question, I'll be happy to vote to re-open.

    • CommentAuthorRyan Budney
    • CommentTimeDec 11th 2010 edited
     
    Willie, modify your statement from "connected" to "path connected" and it's fine. Think of the "Warsaw/Polish circle" standard point-set topology example.

    IMO the question is perhaps more appropriate for math.stackexchange.
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    (a) I prefer to leave the clarification up to the OP, rather than putting words in his mouth.

    Yeah, but I'd even more prefer to have a good/open question to a poor/closed one. I also don't feel like I'm putting words in his mouth. I was really surprised that there was any confusion about the statement of the question.

    (b) I don't see how the question is interesting even in your current re-write.

    I didn't think hard about the question, but it doesn't seem obvious to me. As Ryan points out, discontinuous functions can have connected graphs (e.g. sin(1/x) and 0 if x=0). Certainly any continuous function has a path connected graph. If the graph of a function is path connected, it definitely looks like the function should be continuous, but a proof doesn't spring to mind right now. It's easy to make mistakes about this sort of thing. For example, I've caught people erroneously claiming that a function $\mathbb R\to \mathbb R$ is continuous if and only if its graph is closed.

    I think the question is particularly interesting because the original formulation of the question said that such an example exists. It could be that the formulation I'm suggesting doesn't admit such an example.

    IMO the question is perhaps more appropriate for math.stackexchange.

    I agree that it would be appropriate there, but I don't see why it's inappropriate on MO. So far, the only objections people have raised is that the question itself was unclear. The only other objection I can think of is a vague feeling that "point set topology is for undergraduates", but I don't agree with that one.

    Perhaps the best argument I can give for appropriateness of the question is that it's a well-posed question that I personally want to know the answer to.

    • CommentAuthorWillieWong
    • CommentTimeDec 11th 2010 edited
     

    @Ryan, Anton: thanks for the clarification. After I clicked submit it occurred to me that that intuition only works for path-connected, but no obvious counterexamples came to mind. (A fact that is still quite embarrassing.)

    @Anton: is it possible for a moderator to suggest to the OP this re-write of the question to see if he agrees with it? The fact that it was already modified quite significantly I think means that the OP will be amenable to improving the question. Like I said above, if somehow the question gets put into this improved form, I now agree that it is interesting.

    Also, in view of this discussion, maybe it is more transparent to make the question into two parts

    (a) Can a path connected set be the graph of a discontinuous function?

    If not, then we are done.

    (b) If yes, can the set also be the graph of a continuous function (after a rotation)?

    What do you think?

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    is it possible for a moderator to suggest to the OP this re-write of the question to see if he agrees with it? The fact that it was already modified quite significantly I think means that the OP will be amenable to improving the question. Like I said above, if somehow the question gets put into this improved form, I now agree that it is interesting.

    According to his user page, the OP hasn't been around since his last edit. I'm not patient enough to wait for the OP, I'm (perhaps overly) confident that he wouldn't disagree with the edit, and I'm curious about the answer, so I've just modified the question. I feel like this is a reasonable use of editing powers. Though the modified question shares few of the words of the original, I think it's true to the spirit.

    On another note, some of the comments refer to the original wording of the question. I'm going to clear the ones that no longer make any sense. For posterity, I'll record the comment thread as it stands now:

    Are you sure that claim is valid for all (graphs of) continuous functions? How about a constant function R→R, say x↦0? Unless you rotate it by 90 degrees the graph will be a straight line and hence continuous, whereas if you rotate by 90 degrees it is not the graph of a function at all anymore. I think more details of the question are in order. – Chris Heunen 17 hours ago

    Wouldn't it make more sense to go back and watch the lecture again? Perhaps I'm misunderstanding you but when you say "watch" it indicates to me this was a recorded lecture. As is, it's not clear to me what you're asking about. – Ryan Budney 17 hours ago

    This sounds similar to taking the inverse of an injective bounded operator with dense range - this is usually not continuous and only densely defined. The graphs are rotations of one another, but it is not classical analysis.. – Ollie Margetts 16 hours ago

    What was the role of this particular paper in Wilf's talk? Was it meant to be an example of bad exposition? – Scott Carnahan♦ 13 hours ago

    How do all the above comments relation to the question? (There is currently no claim in the question txt, there word 'watch' does not appear in it, Wilf is not mentioned...) If the question was edited, an effort to render exiting comments not nonsensical would be helpful! – Mariano Suárez-Alvarez 11 hours ago

    @Mariano: the question has been edited in a way that make the initial comments appear nonsensical. The current formulation has various trivial answers so it seems rather poorly phrased. Also, the attributions of Herbert Wilf and Thurston appear to be kind of random. – Ryan Budney 10 hours ago

    Even now the question is poor: take f(x)=x on [0,1) and x+1 on [1,2]. Then f is discontinuous and f−1 is continuous (on their domains, of course). If we assume the set S to be compact, then every function arising this way is continuous and the question is vacuous again. We can, probably, try to salvage it in some way but I prefer to leave that to OP. – fedja 9 hours ago

    meta.mathoverflow.net/discussion/828/… – Anton Geraschenko 4 hours ago

    • CommentAuthorRyan Budney
    • CommentTimeDec 11th 2010 edited
     
    I think the question has an easy answer, and it's essentially a 1st course in analysis type question that reduces pretty quickly to an intermediate value theorem application. Here's a more general statement. Let $f : R \to R$ be continuous and $h : R^2 \to R^2$ a homeomorphism. Then if $h(graph(f))$ is the graph of a function $g : R \to R$, then $g$ is continuous.

    Sketch: consider the function $P : R \to R$ given by $P(x) = projection onto x-axis of h(x,f(x))$. This is a continuous monotone function by the assumptions, so it is an open map by the intermediate value theorem, so it's inverse exists and is continuous. The claimed function $g$ is then $g(x)=f(P^{-1}(x))$.

    The question gives me the feeling it's a homework problem, but I've never seen it before.
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    @Ryan: Yes, I think that works, except the claimed function $g$ should be the projection onto the y-axis of $h((P^{-1}(x),f(P^{-1}(x))$, not $f(P^{-1}(x))$. The argument that $P$ is open is the crux ... it's slightly different from the intermediate value theorem, but it's certainly elementary. Thanks.

    For what it's worth, I still think it's worth reopening the question. It's true that this would make a very nice homework question.

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    I just chatted with Scott, who also "voted" to reopen, so I've reopened the question.

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    Perhaps stating the obvious, I think something got lost on the way from the OP's hearing of Herbert Wilf's talk and the original formulation of the question. I upvoted Pietro's answer because I think he has a finger on what the issue might have been: in a bygone age before the notion of function had been clearly recognized and formulated, mathematicians worried that rotating a continuous sawtooth through 45 degrees would produce a function which is discontinuous according to Cauchy's definition of continuity. This is according to Proofs and Refutations, top of page 130 (see also footnote 2). Perhaps Wilf was alluding to this sort of thing.

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