Let $I=[0,1]$ and $I\times I$ be a subspace of $\Bbb R^2$.
Find an example satisfying the following conditions:

  • $A$ and $B$ are subsets of $I\times I$.
  • $A$ and $B$ are connected.
  • $(0,0), (1,1)\in A$ and $(0,1), (1,0)\in B$.
  • $A \cap B=\varnothing$.
  • $\begingroup$ Could you tell us something about your thoughts on this problem? Beyond the obvious that $A$ and $B$ can't both be path connected. $\endgroup$
    – kahen
    Commented Oct 3, 2013 at 6:24
  • $\begingroup$ @kahen Obvisouly $A,B$ cannot be both simultaneously open, but they cannot be open, since then they would be path connected. $\endgroup$
    – Pedro
    Commented Oct 3, 2013 at 6:25
  • $\begingroup$ @PedroTamaroff I was thinking more along the lines that some variant of the intermediate value theorem forces $A$ and $B$ to intersect if they're both the images of continuous curves. $\endgroup$
    – kahen
    Commented Oct 3, 2013 at 6:28
  • $\begingroup$ I studied the concept - connected - recently, but I don't know "path connected". It is possible to solve this problem if we use the topologist's sine curve, I think. $\endgroup$
    – Hobin. J
    Commented Oct 3, 2013 at 6:33
  • 1
    $\begingroup$ @copper.hat Somebody had $\endgroup$ Commented Oct 3, 2013 at 20:23

2 Answers 2


The standard example of a set which is connected but not path connected is $(x, \sin(1/x))$ for $0 < x < 1$ together with a second piece such as $(0,y), -1 \le y \le 1.$ You can use this to solve the problem. Set $A$ to be the union of two sine curves $\sin(1/x)$ rotated $90$ degrees together with one extra point. In more detail, $(x, \sin(1/x))$ with $0 < x \le 1$ has a "tail" at $(1, \sin(1))$ and approaches a "wall" at $(0, y), -1 \le y \le y$. Rotate this curve so its tail is at $(0,0)$ and its wall is the the line from $(1,0)$ to $(0,1).$ Make a second copy whose tail is at $(1,1)$ and whose wall is again the line from $(1,0)$ to $(0,1)$, but coming from the other side. Their union is not connected without adding a piece of their common wall, so add in the point $(1/2, 1/2)$. The union of those three pieces is connected. Take $B$ to be everything else: $B = I \times I - A$.

To show that $B$ is connected, let $L$ be the wall, the line from $(0,1)$ to $(1,0)$. Points in $I\times I - L$ that are below and to the right (BR) of the sine curves can be connected with a path. And similarly for points above and to the left (UL). So if there is a disconnection of $B$ as $ U \cup V$, then one of them (say $U$) contains BR, and $V$ contains UL. What about a point $p \in L - (1/2,1/2)$? It must be in either $U$ or $V$, but every neighborhood of $p$ contains points of BR and UL,and so contains points of $U$ and $V$, contradicting $U \cap V = \emptyset$

  • 3
    $\begingroup$ If set $A$ is the line from $(0,0)$ to $(1,1)$, its complement (in the square) is the disjoint union of two nonempty open subsets of the square, one of which contains $(0,1)$, and the other $(1,0)$. Then you can't have a connected set containing both points disjoint from $A$. $\endgroup$ Commented Oct 3, 2013 at 18:15
  • $\begingroup$ Oh right, I guess my solution is bogus! $\endgroup$ Commented Oct 3, 2013 at 18:17
  • $\begingroup$ Thanks for pointing out my bug. You need to add one point of the common wall to make two sine curves connected. Then you can take the other set to be the complement. $\endgroup$ Commented Oct 3, 2013 at 20:08
  • 2
    $\begingroup$ Now what's missing is the argument for the connectedness of $B$. Or at least a hint. It's not exactly obvious that $B$ is connected. But good thinking on the bug-fix, that alone deserves an upvote. $\endgroup$ Commented Oct 3, 2013 at 20:15
  • $\begingroup$ That reminds me of an answer I once wrote. The idea is similar and $B$ seems to be connected, as the space "below" the sine curve approaches the line from $(0,1)$ to $(1/2,1/2)$. $\endgroup$ Commented Oct 3, 2013 at 20:48

David Goldberg was definitely along the right track. We can complete the idea as follows. (Edit: the first answer I gave was written in intuitive language, with the idea that the formal mathematical details would be routine to fill in. After reading a comment below, I decided to completely rewrite the answer, so as to give a more formal, precise description.)

Define a connected set $P = P_1 \cup P_2 \cup P_3 \cup P_4 \cup P_5$ where

$$P_1 = \{(x, \sin(1/x): 0 < x \leq 1\}$$

$$P_2 = \{(t + 2(1-t), t\sin(1)): 0 \leq t \leq 1\}$$

$$P_3 = \{(0, 0)\}$$

$$P_4 = \{(x, \sin(1/x): -1 \leq x < 0\}$$

$$P_5 = \{(-t - 2(1-t), t\sin(-1)): 0 \leq t \leq 1\}.$$

Here $P_1$ and $P_4$ are right-hand and left-hand branches of a topologists' sine curve, $P_2$ and $P_5$ are line segments which connect these branches to the points $(2, 0)$ and $(-2, 0)$ respectively, and $P_3$ is just some accumulation point along the $y$-axis that we adjoin so that $P$ is connected. The choice $P_3 = \{(0, 0)\}$ is thus somewhat arbitrary.

Somewhat similarly, define a connected set $Q = Q_1 \cup Q_2 \cup Q_3 \cup Q_4 \cup Q_5$ where

$$Q_1 = \{(x, \sin(1/x) + 1/4): 0 < x \leq 1\}$$

$$Q_2 = \{(t, t(\sin(1)+1/4) + 2(1-t)): 0 \leq t \leq 1\}$$

$$Q_3 = \{(0, 1/2)\}$$

$$Q_4 = \{(x, \sin(1/x) - 1/4): -1 \leq x < 0\}$$

$$Q_5 = \{(-t, t(\sin(-1)-1/4) - 2(1-t)): 0 \leq t \leq 1\}$$

Here $Q_1$ is a right-hand branch of the topologists' sine curve but translated up by $1/4$, which has accumulation points $(0, t)$ with $-3/4 \leq t \leq 5/4$. Similarly, $Q_4$ is a left-hand branch translated down by $1/4$, which has accumulation points $(0, t)$ with $-5/4 \leq t \leq 3/4$. Clearly the point of $Q_3$ is an accumulation point of each branch. Finally, $Q_2$ and $Q_5$ are line segments which connect these branches to the points $(0, 2)$ and $(0, -2)$, respectively. The set $Q$ is connected for the same reason $P$ is connected.

Readers may wish to draw a picture. It is clear by construction that $P$ and $Q$ do not intersect (particularly, the topologist-sine-curve branches do not intersect because they are vertical translates of each other).

Finally, we apply a homeomorphism which will map the four diamond points $(2, 0)$, $(-2, 0)$, $(0, 2)$ and $(0, -2)$ to the four corners of $I \times I$, and map $P \cup Q$ into $I \times I$. For example, we may first apply a rotation which sends the diamond points to the four points $(\pm \sqrt{2}, \pm \sqrt{2})$, followed by scalar multiplication by $1/2\sqrt{2}$ to map to the four points $(\pm 1/2, \pm 1/2)$, followed by a translation $(x, y) \mapsto (x + 1/2, y + 1/2)$.

  • $\begingroup$ Why the "tail" of T_0 is (2, 0)? We have the subspace I×I of R^2, so (2,0) is not well-defined? $\endgroup$
    – Hobin. J
    Commented Oct 4, 2013 at 0:19
  • $\begingroup$ At the end, I said we rotate, dilate, and translate so as to meet the original conditions of the OP. (I chose to pitch my answer at an intuitive level, trusting that such details were completely routine. But I am happy to edit in a more formal description, so that nothing is left to chance.) $\endgroup$
    – user43208
    Commented Oct 4, 2013 at 0:41
  • $\begingroup$ Could you draw the gragh of P and Q? I can't imagine this situation. $\endgroup$
    – Hobin. J
    Commented Oct 4, 2013 at 0:54
  • $\begingroup$ Sorry, I'm no good at dealing with graphics. But I hope the description I've just edited in will be clear. I trust you've seen this topologist sine-curve business? $\endgroup$
    – user43208
    Commented Oct 4, 2013 at 1:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .