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Show that any set obtained by removing a single point from $\mathbb{R}^2$ is still connected, where $\mathbb{R}$ is the real numbers.

Then show that $\Bbb H = \{(x,y) : x>0\}$ is connected. By considering the function $$f(x, y)/x,$$ or otherwise, show that there are precisely two continuous functions $f : \Bbb H \to \Bbb R$ such that $$f(x, y)^2 = x^2$$ for all $(x, y) \in \Bbb H$.

This is a problem I saw yesterday and it's quite interesting, but I'm not having much luck with solving it! Can anyone help out with a proof?

dfeuer
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Tom
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1 Answers1

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HINT: Let $p\in\Bbb R^2$, and let $X=\Bbb R^2\setminus\{p\}$. If $x$ and $y$ are distinct points of $X$ either the line segment $\overline{xy}$ lies entirely in $X$, or $p$ is on that line segment. Show that in the latter case there is a path from $x$ to $y$ consisting of two line segments that both lie entirely in $X$. Finally, show that if $X$ were not connected, there would be two points of $X$ with no path in $X$ connecting them.

Brian M. Scott
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  • thanks! that makes sense I think. what about the other bits? – Tom Oct 28 '13 at 07:34
  • @Tom, if I read the other bits correctly, they are suggesting extremely awkward ways to answer the question that may nevertheless be useful for something else. Do you agree, Prof. Scott? – dfeuer Oct 28 '13 at 07:38
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    @Tom: You’re welcome. If $p,q\in\Bbb H$, then $\overline{pq}\subseteq\Bbb H$, so connectedness of $\Bbb H$ is no problem. A hint for the rest: $x^2=y^2$ if and only if $x=\pm y$. This should tell you what the two functions have to be, and the hint in the question then provides a way to show that they’re the only ones. – Brian M. Scott Oct 28 '13 at 07:40
  • @dfeuer: Looking at $\frac{f(x,y)}x$ seems a pretty reasonable way to show that the two functions (once you’ve found them) are the only continuous ones. – Brian M. Scott Oct 28 '13 at 07:44
  • @BrianM.Scott, maybe I misread. I thought that proving there are exactly two functions was intended to be a lemma to the connectedness proof, which struck me as twisted. – dfeuer Oct 28 '13 at 07:45
  • @dfeuer: Other way round, I think: connectedness of $\Bbb H$ is to be used to show that the quotient must be constant. – Brian M. Scott Oct 28 '13 at 07:51
  • thanks - I'm not sure how to do this bit still - would you mind giving a proof for the question? Also, can it be done without using path connectedness? – Tom Oct 28 '13 at 08:03
  • @Tom: Which bit? I’ve not thought about alternatives, because path connectedness is by far the easiest way to go. If you’re having trouble getting from there to connectedness, remember that $X$ is connected iff every continuous function $f:X\to{0,1}$ is constant, where ${0,1}$ has the discrete topology. – Brian M. Scott Oct 28 '13 at 08:14
  • I've got now that path connected implies connected but I can't see how to do the final bit about there being two funtions satisfying the condition – Tom Oct 28 '13 at 08:16
  • @Tom: Which part? Finding two continuous functions that satisfy it, or showing that they’re the only ones. The first part comes from the hint in my earlier comment; the second, from the connectedness of $\Bbb H$. I’m willing to say more, but I’d like to know just where you’re hung up. – Brian M. Scott Oct 28 '13 at 08:18
  • I think that f(x,y) can clearly be equal to x or -x? But then I'm not sure how to show these are the only ones – Tom Oct 28 '13 at 08:20
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    @Tom: That’s right. More generally, we must have $f(x,y)=\pm x$ at every point, whether or not $f$ is continuous. Now look at the function $g(x,y)=\frac{f(x,y)}x$; if $f$ is continuous, so is $g$. What values can $g$ assume? – Brian M. Scott Oct 28 '13 at 08:25
  • so g can only take values 1 or -1, but g is continuous and hence we must have g=1 for all x>0 or g=-1 fpr all x>0. Is that right? – Tom Oct 28 '13 at 08:32
  • @Tom: That’s exactly right. (I like the argument: it’s quite slick.) – Brian M. Scott Oct 28 '13 at 08:32
  • It's really neat I agree. could we generalise the argument to see how many functions there would be from the real plane to the reals such that $$f(x,y)^2 = x^2$$? – Tom Oct 28 '13 at 08:35
  • @Tom: Yes, if you’re careful. It’s not two any longer. – Brian M. Scott Oct 28 '13 at 08:37
  • since the line x=0 origin is included in the real plane, then f(0,y) = 0 on this line. But as before for x>0 and x<0, we must have f(x,y) = 1 or -1. But then f cannot be continuous so are there no such functions? – Tom Oct 28 '13 at 08:40
  • @Tom: The line $x=0$ is indeed what makes this different, but not in that way: the function $g$ that I defined before has to be continuous and therefore constant on the left and right open half-planes, but since it’s not defined on the $y$-axis, it can change sign there. There are actually more than two possibilities, though still only a small finite number; can you see what they are? – Brian M. Scott Oct 28 '13 at 08:43
  • Got it - so there are four possibilities for the possible combinations of f(x,y) being 1 or -1 for x>0 / x<0 – Tom Oct 28 '13 at 08:46
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    @Tom: Exactly. You can think of them very geometrically, too. The $1/1$ map just squashes the plane to the $x$-axis. The $-1/-1$ map reflects it in the $y$-axis and then squashes it to the $x$-axis. The other two maps fold it over the $y$-axis, either to the left or to the right, and then squash it. – Brian M. Scott Oct 28 '13 at 08:49