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Let $f$ be a continuous real function defined on the unit square $[0,1]\times[0,1]$, such that $f(x,0)=-1$ and $f(x,1)=1$.

If we walk from $y=0$ to $y=1$ on any path, we must cross some point where $f=0$. Intuitively, there is a connected "wall" (a curve) that separates $y=0$ from $y=1$, on which $f=0$, like this:

enter image description here

My question is: is there a theorem that formalizes this intuition (if it is correct)? In particular: is there a theorem which implies that there exists a connected line that goes from $x=0$ to $x=1$, on which $f=0$, as in the picture?

Erel Segal-Halevi
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  • My guess is that it would just be an extension of the IVT to higher-dimensions. – Sean Henderson May 13 '15 at 19:30
  • First of all, if $f$ is not injective, there may be more than place $f=0$ for a given $x$. For example, $f(x,y)=cos(3\pi y)$. – ET93 May 14 '15 at 02:45
  • @ET93 OK, but does there always exists a connected path from $x=0$ to $x=1$ on which $f=0$? – Erel Segal-Halevi May 14 '15 at 03:02
  • This is more related to the implicit function theorem than the IVT. If you assume the conditions of the IFT, you can get a differentiable function from some open set of $[0,1]$ to the points where $f=0$. – ET93 May 14 '15 at 03:36
  • Suggested plan for proof - assume it's not true - ie, the set of zeros is not connected between x=0 and x=1. Then there are 2 disjoint open sets, one containing the x=0 side, and the other one containing the x=1 side, which cover the entire set of zeros. Now prove that in the remaining piece of the square not covered by either open set there is one connected component touching both y=0 and y=1. Again, otherwise there would be 2 disjoint open sets, one containing y=0, the other one containing y=1, which cover the remaining part of the square, and I expect that this is not possible – ivancho May 14 '15 at 03:55
  • One theorem you might one to look at is the Poincare-Miranda theorem. It is not exactly what you asked for, but it does formalize that 'intuition'. – Dongryul Kim May 14 '15 at 10:51
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    You shouldn't say "connected line" because the connected set you want may well be like the topologist's sine curve at the left and right sides of the square, so there is no notion of line that would capture what you want. Instead you could say "connected set containing at least one point on each of the left and right sides of the square". – user21820 May 15 '15 at 05:37
  • @DongryulKim Actually this theorem answers another question which I asked some time ago: http://math.stackexchange.com/questions/1070472/is-this-two-dimensional-version-of-the-intermediate-value-theorem-correct?rq=1 Would you like to write an answer to that question? – Erel Segal-Halevi May 15 '15 at 08:05

3 Answers3

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EDIT 2: The question has been changed to ask not for a curve, but for a connected set. The below no longer applies but I am leaving it here in case others find it useful.

EDIT: The original version of the question was unclear; now that it has been clarified that one seeks a curve along which $f=0$, I believe the claim is false.

First, it is obvious that

1) The sub- and super-level sets where $f<0$ and $f>0$, respectively, are separated, and

2) The level set $f^{-1}(0)$ does not have to be a curve (consider for instance $$f(x,y) = \begin{cases}0, &\frac{1}{4} < y < \frac{3}{4}\\4y-3, &y\geq\frac{3}{4}\\4y-1, &y \leq \frac{1}{4}.\end{cases}$$

Now your question (if I'm interpreting it correctly) is whether the zero level set contains some continuous curve. I don't think this is necessarily the case. Consider for instance the set $S$ given by two copies of the topologist's sine curve: $$S = \left\{ \left( x,\frac{1}{2}+\frac{1}{2}\sin\frac{1}{2x-1}\right)\ \Big\vert\ x \neq \frac{1}{2}\right\} \cup \left\{\left(\frac{1}{2},y\right)\ \Big\vert\ 0 \leq y \leq 1\right\}.$$

enter image description here

$S$ is closed and so the distance function $d(x,y)$ of $(x,y)$ to $S$ is well-defined and continuous. Also define a sign function $\epsilon(x,y)$ to be 1 above and -1 below the sine curve, i.e.

$$\epsilon(x,y) = \begin{cases}\textrm{sign}\left(y-\frac{1}{2}-\frac{1}{2}\sin\frac{1}{2x-1}\right), &x\neq \frac{1}{2}\\0, & x = \frac{1}{2}\end{cases}$$ and set $f=\epsilon d$ to get a counterexample where $f^{-1}(0)$ is not path-connected.

(This $f$ does not exactly satisfy your boundary conditions at $y=\pm 1$, but this could be fixed by e.g. taking $f(x,y) = (\epsilon d)(x,[4y-1]/2)$ on $y\in (1/4,3/4)$ and extending appropriately.)

user7530
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    OK, but how does this prove that there exists a connected path from $x=0$ to $x=1$ on which $f=0$? – Erel Segal-Halevi May 14 '15 at 03:03
  • @ErelSegal-Halevi I've edited my answer. – user7530 May 14 '15 at 04:21
  • I think I see what you mean: $f$ changes continuously from being negative below the sine to being positive above the sine; it is 0 only on the sine, which is not path-connected. But, the sine is connected, so this still isn't a counter-example for my question. – Erel Segal-Halevi May 14 '15 at 04:58
  • Then you need to state, very precisely, what you mean in your question. There is not "continuous 'wall' (a curve) that separates y=0 from y=1, on which f=0" in the example above. – user7530 May 14 '15 at 04:59
  • You are right; I fixed the question to say "connected". – Erel Segal-Halevi May 14 '15 at 05:00
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    @ErelSegal-Halevi I think what you want to ask is, "does $f^{-1}(0)$ have a connected component that includes a point on both $x=0$ and $x=1$." The revised question still asks for a "curve" which does not exist in the example above. – user7530 May 14 '15 at 05:07
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    Yes, I meant to ask about a connected component. Does the word "curve" implies path-connectedness? – Erel Segal-Halevi May 14 '15 at 05:22
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    Of course the precise definition is convention dependent but at minimum a curve is a continuous map of a real interval. – user7530 May 14 '15 at 05:42
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Yes, there is a theorem from Topology (cf. `Topology,' by James R. Munkres) that says if $X$ and $Y$ are topological spaces, and if $f:X\to Y$ is continuous, and if $X$ is connected, then $f(X)$ is connected. Furthermore, if $A$ is a connected subset of the real number line, then $A$ must be an interval. To apply it to your case: The image of a path from $y=0$ to $y=1$ is a connected subset of the square. Then apply the previous results.

treble
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    I am not sure I understand. How does this prove that the line on which $f=0$ is indeed connected (like a wall; see the picture I added)? – Erel Segal-Halevi May 14 '15 at 02:39
  • I am not sure I understand. How does this prove that there exists a connected path from $x=0$ to $x=1$ on which $f=0$? – Erel Segal-Halevi May 14 '15 at 03:03
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    @ErelSegal-Halevi Your question really wasn't clear, as user7530 indicated. Your question really isn't about the IVT per se, as you first indicated. It is about the structure of the level set of a continuous function. – treble May 14 '15 at 07:28
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Converting my comment to a loose idea of proof:

Assume it's not true - ie, the set of zeros of f is not connected between x=0 and x=1. Then there are 2 disjoint open sets, A and B, one containing the x=0 side, and the other one containing the x=1 side, which cover the entire set of zeros. Now let's prove that in the remaining piece of the square not covered by either A or B there is a connected component touching both y=0 and y=1. This would contain a path where f is non-zero (well, not exactly a path, but it would kind of break the notion of 'every connection between the two sides must equal zero somewhere'). Again, let's assume otherwise, so there are 2 disjoint open sets, C and D, one containing y=0, the other one containing y=1, which cover the remaining part of the square.

Now we prove that the boundaries of A and C, $\partial A$ and $\partial C$ intersect - with some hand-waving, the connected component of A containing the entire x=0 side has points both in the interior of C: (0, 0), and points in the interior of the complement of C: (0, 1), and since it's connected, it must also have points on the boundary of C, and the limit points of these will still be on the boundary of C, but will also be on the boundary of A. Now look at a single point $x \in \partial A \cap \partial C$. It is not inside A or C, since they are open, but it also cannot be in B or D. For example, if we assume $x \in B$ then there is an open neighborhood of x fully contained in B, and therefore disjoint from A, which is a contradiction with $x \in \partial A$. So x is not in $A \cup B \cup C \cup D$, which is a contradiction, we assumed that those 4 sets cover the square.

ivancho
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    You misapplied the finite intersection property, and the resulting claim is incorrect. It is easy to find 4 closed sets covering a square such that any 3 have non-empty intersection but all 4 have empty intersection. – user21820 May 15 '15 at 02:29
  • Sorry I should say 4 closed sets in a square, not "covering a square". – user21820 May 15 '15 at 02:35
  • You are absolutely right, finite intersection is not the correct way to prove the statement. – ivancho May 15 '15 at 02:40