Help proving that this metric space is not path connected

Question

Consider the metric space embedded in $S^1$ with the intrinsic metric(the distance between two points is the length of the shortest arc connecting them):

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Notice there are $3$ 'gaps' in the circumference of the circle. We can fill each of these gaps with a smaller copy of this space, and continue like this recursively, below is a diagram of the second iteration in this process:

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And the third iteration:

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If we keep doing this on each, newly created gap, we obtain an ascending sequence of spaces, and we can thus take the union of all finite iterations such as the ones above. Call this space $\mathcal{U}$ (Which will look like a fractal). I am interested in properties of this space, which would look something like this (excuse the imprecision)

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Question: Is the space $\mathcal{U}$ path connected? Although it seems intuitively clear that the space $\mathcal{U}$ is not path connected, how does one formally prove it? The intuitive argument is that if you want to travel from $0$ to $1$, you need to initially stick to the left-segment, then when you get to the end of this left segment, and the start of the next circle, again to traverse this circle, you would need to stick to the left segment, and so you're forced to take the left-most segment each time, but the "path" that originates if you try to do this has a single point missing precisely in the middle, and so there is no path.

Answer 1

Denote by $U_n$ the set of points at iteration $n$, so that $\mathcal U=\bigcup_{n=1}^\infty U_n$ is the space under consideration.

Now define an equivalence relation $\sim$ on $\mathcal U$ where $x\sim y$ if for some $n$, we have $x,y\in U_n$, and $x$ and $y$ are in the same connected component of $U_n$. Note that $x\sim y$ if and only if $x$ and $y$ are in the same component of the first finite iteration $U_n$ containing them both, i.e., successive iterations do not connect points that were not already connected. Denote by $[x]$ the equivalence class of $x$ under $\sim$.

Now suppose $x\not\sim y$, and $x,y\in U_n$. Then we have $$d([x],y)\geq d([x],[y]\cap U_n)\geq \frac{1}{2\cdot 3^n},$$ (in your picture, $[x]$ will never, at any stage, cross the midway distance on a dotted line at stage $n$ separating $x$ and $y$).

Therefore $y\notin \overline{[x]}$, and so the equivalence classes of $\sim$ are closed.

But now suppose $\gamma\colon [0,1]\to \mathcal U$ is continuous. There are countably many equivalence classes of $\sim$ (any point is in the equivalence class of one of the finitely many components of some $U_n$), call them $[x_n]$, so we have $$[0,1]=\bigcup_{n=1}^\infty \gamma^{-1}([x_n]).$$

By continuity this is a countable union of disjoint closed sets, so by a theorem of Sierpiński we have $[0,1]=\gamma^{-1}([x_n])$ for some $n$. Thus two points may be connected with a continuous path if and only if they are $\sim$-equivalent. Since points in distinct components of each $U_n$ are not equivalent, $\mathcal U$ is not path connected.

Remark

I don't know for sure if this space is even connected. I suspect it is not, but a rigorous proof of that escaped me, and I liked the Sierpiński argument so I thought I would share, since it does answer the question asked about path connectedness.

Answer 2

I put this in a comment but the picture wouldn't properly render unless you click on it, so putting it here in an attempt at an answer that I may come back to later.

I admit I am not using intrinsic metric, but topologically (or even metrically?) it shouldn't matter. When you scale a circle down, it seems from the picture (inscribe a hexagon) that the diameter and the circumference decrease by a factor of $2$, not $3$ (though OP insist it is $3$, and I should take a more careful look). One could represent the set of centers of the circles, start at the origin, fix three vectors $u,v,w$ of equal length making $120^\circ$ angles pairwise (e.g. (minus) cube roots of $1$ in the complex plane), and consider finite sums $\displaystyle e_1+\frac{e_2}2+\frac{e_3}4+...+\frac{e_k}{2^{k−1}}$ where $e_i\in\{u,v,w\}$. Each such finite sum being a center, the radius decreases accordingly, three circular arcs are taken, right, top-left, bottom-left. (In the first picture further down, $u$ not shown but it goes from the center of the large circle to the center of the smaller circle on the left from the next iteration. The confusing "(minus)" above would disappear if one reflects the whole picture about the $y$-axis (which was done in the second picture further down)). In each connected triple of red line segments, the common point is the center of a circle, each of the three red segments being a radius, and three (closed) circular arcs from this circle go into the set, the other three (open) circular arcs do not (and I believe none of their points does, even at later stages, need to take a look)).

Here is a picture where your set does not intersect the union of the red line segments, and which I believe could be used to make your intuitive argument precise.

It would also technically help to consider topological arcs instead of paths, where a (topological) arc is a path that has no self-intersections (never goes back to a point already visited). It is known that if two points are connected by a path, then they are also connected by a (topological) arc.

(Regarding a comment by Alex Ravsky which is a bit short so I may not follow what was meant to be said there, but if I do then: Aren't endpoints of circular arcs in the top-left path-component lying on a presumed "separating" line? Edit. I did reread that comment, it is about the construction with two gaps, instead of three, a different matter than what I was assuming, just looking at the pictures that go with three gaps. With two gaps indeed there would be a separating line (though I feel I need to see a picture to make sure how exactly the smaller arcs, that are added at each step, are joined to bigger arcs, e.g. at an endpoint or a midpoint of a smaller arcs). One could see some "trees" though in the construction with three gaps and perhaps try use them.)

Regarding the answer that is already posted I believe it is correct though I need to read it more carefully to understand some of the details. I tend to believe that the set is connected, even if not path-connected (this belief needs a proof, of course).

Edit. I read the comments by OP posted below, they are a bit over my head (not enough background on my part in algebraic/geometric topology, or differential geometry, whichever this belongs to) though perhaps I get some of the ideas, thank you. I seem to be interested in a different problem that also makes sense to me, like assuming these were honesty good pictures of a subset of the plane with the usual metric (that is what I took it to be on quick reading first). In this case the version with three gaps (instead of only two) seems just about right, an interesting problem. I may try ... to use some software to make a sketch in the plane (whether with a factor of 2 which I got used to, while staring at the pictures) or a factor of 3 (about which I need to think).

For what it is worth, here is a more precise picture of what I was "seeing" while looking at the pictures posted by OP. What is below is an honesty good embedding in the plane (with the usual metric inherited from the plane), that seems to parallel the picture posted by OP. First four levels are shown, along with a (dashed) line which "almost" appears to be separating, except that many endpoints of circular arcs of the "top-right" path-component lie on this line. (I reflected the previous picture (above, in this answer) about the $y$-axis since in my mind I think of certain finite sums involving ($\frac1{2^k}$ multiples of) vectors $u,v,w$ that correspond to the cube roots of $1$ in the complex plane. Each such finite sum is the center of a circle with appropriate radius.) Obviously the picture below closely parallels the construction by OP and perhaps could be employed in some useful way. (E.g. it looks like one could use it to illustrate the proof that the space in question is not path-connected.)

(Credit to Ivan Johansen and his program "Graph" available at https://www.padowan.dk/ , used to make the above picture.)