MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi all. Can you help me with this? I have a square $S$ in euclidean plane with edges $A,B,C,D$ and a closed set $F$ in $S$ such that $F\cap A=F\cap C=\emptyset$, and $F\cap B$ and $F\cap D$ are nonempty. Assume that any curve in $S$ starting on $C$ and ending on $A$ intersects $F$. Does it follow that there exists a curve in $F$ starting on $B$ and ending on $D$?

Intuitively, it seems trivial, but I don't know how to proove it formally. Thanks a lot, Peter

share|cite|improve this question
thank for answers. If I change it to "does F has to have a connected component connecting B and D", is it true? Peter – Peter Franek Nov 5 '10 at 21:18

No, take $S=[-2,2]\times[-2,2]$ and $F$ the closure of the graph of $\[-2,2]\setminus\{0\}\ni x\mapsto \sin(1/x)$.


However, it is true that there is a connected component of $F$ that meets both the (closed) edges $B$ and $D.$ Equivalently, there is a connected component of the set $G:=F\cup B\cup D$ that contains both $B$ and $D.$ Suppose not, by contradiction. Then, the same holds for some open neighborhood $V$ of $G$ in $S$ (this is easily shown recalling that a nested intersection of connected compacta is a connected compact). This amounts to saying that the edges $B$ and $D$ have disjoint open nbds, resp, $U_B$and $U_D$ whose union is $V$. The assumption that $S\setminus F\\ $ contains no paths connecting $A$ and $B$ implies that the edges $A$ and $B$ also have disjoint open nbd $U_A$ resp., $U_C$ whose union is $S\setminus F\\ .$ In conclusion, we have covered the square with four nbds of the edges, in such a way that nbds of opposite edges do not meet. This leads to a contradiction. Consider a partition of unity subordinate to the covering $\{U_A, U_B, U_C, U_D\},$ and use it to define a vector field $X$ on $S$ such that on $U_i\cap U_j$ the field $F$ is a convex combination of the exterior normal to $U_i$ and the exterior normal to $U_j$. This implies that $X$ has topological degree 1 zero, and never vanishes, a contradiction. Otherwise, you may use $X$ to retract the square on its boundary, another contradiction.

share|cite|improve this answer
Yes, thats right. What would be sufficient for me, is: "does F have to have a connected component, connecting a point on B and a point on D?" Do you think now it holds? Thanks a lot. – Peter Franek Nov 5 '10 at 21:17
Yes, that is true. I do not have a reference handy, but in case I'll add a proof. – Pietro Majer Nov 5 '10 at 21:22
Thanks. If its easy, could you give me some hint? Peter – Peter Franek Nov 5 '10 at 21:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.