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It's the union of a sequence of interlocked "chains" formed from closed semicircles. The chains can be seen having different shades of gray in the picture.

There's no line in the middle, just the endpoints of arcs which form the semicircles, else it would be trivial.

Connected means it can't be decomposed into 2 nonempty open sets.

I think it's not connected, so I thought about taking one semicircle and including all semicircles which intersect small neighborhoods of the original semicircle's endpoints, iterating this process I can get an open subset which is not the entire thing (if the neighborhoods are chosen sufficiently smaller and smaller). But I would need it also to be closed in order to separate it, but I don't see that it has to be.

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  • $\begingroup$ I'd be interested to know what subset of $\Bbb R$ is homeomorphic to the set of all endpoints of semicircles. It's really not clear from the picture, and I think that it might be the crucial part of the matter. $\endgroup$ – MJD Oct 11 '14 at 19:02
  • $\begingroup$ @MJD Well, it's a countable dense linearly ordered set with a minimal and maximal element. Shouldn't that automatically be the same as the rational numbers in [0,1]? $\endgroup$ – user2345215 Oct 11 '14 at 19:06
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    $\begingroup$ Do you have a definition for this sequence of interlocked chains? I think your hunch is correct, and depending on the definition I think a proof by induction could do the trick $\endgroup$ – graydad Oct 11 '14 at 19:09
  • $\begingroup$ @graydad It would be hard to define it precisely, it should be easy to see from the picture. Basically, you can imagine it extends infinitely to left and right. Now start in some place on the middle line and start moving in one direction. If you see endpoints of 2 differently oriented circles, you traverse it using 2 circles, if the endpoints are oriented in the same way, you use 3 circles to go through. $\endgroup$ – user2345215 Oct 11 '14 at 19:13
  • $\begingroup$ if I am not missing something, it seems that by construction the centers of all circles are excluded from the line. $\endgroup$ – Wolphram jonny Oct 11 '14 at 19:17
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This set, which I will denote $S$, is not connected, no matter how you define it (as long as the closed arcs are pairwise disjoint), here is a proof.

Form an equivalence relation $\sim$ on $R^2$ with equivalence classes being the closed arcs you drew and singletons not contained in such arcs.

Then take $X=R^2/\sim$ with the quotient topology, $q: R^2\to X$ is the quotient map. Since every equivalence class is a closed subset of $R^2$, the space $X$ is $T_1$. Then it is easy check that the decomposition of $R^2$ we defined is "upper semicontinuous". This implies that the quotient space is normal (since $R^2$ is normal). See page 8-10 in Daverman's book for definitions and proofs. Incidentally, our space $X$ is homeomorphic to $R^2$, this is a special case of Moore's theorem. We will not need this, however.

Let $T=q(S)\subset X$ denote the projection of $S$. Then $T$ is countable and infinite (this is where I use the fact that all closed arcs are pairwise disjoint). Since $X$ is $T_4$, you can separate any two points $t_1, t_2\in T$ by a continuous function $f: X\to {\mathbb R}$: $f(t_1)\ne f(t_2)$. Since $f(T)$ is countable, we can find a point $r\in {\mathbb R}$ between $f(t_1), f(t_2)$, which is not in $f(T)$. Then $U_1=(fq)^{-1}((-\infty, r))$, $U_2=(fq)^{-1}((r, \infty))$ are disjoint open subsets of $R^2$ separating $q^{-1}(t_1), q^{-1}(t_2)$. The sets $U_1\cap S, U_2\cap S$ are disjoint and their union is the entire $S$. Therefore, we just proved that any two points in $S$ which are not on the same arc, belong to distinct connected components of $S$. In particular, $S$ is disconnected and connected components are the circular arcs.

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  • $\begingroup$ Very nice! I was actually verifying the upper semicontinuity by constructing the open set in my question, but I didn't notice it. Thank you very much! $\endgroup$ – user2345215 Oct 11 '14 at 23:34

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