Countable number of arcs between two intervals

Question

First consider the two intervals $I_0=\{0\}\times [0,1]$ and $I_1=\{1\}\times [0,1]$ in the plane.

Suppose that, for each $n\in\omega=\{0,1,2,...\}$, $A_n$ is an arc (a homeomorphic image of $[0,1]$) contained in $\mathbb [0,1]^2 \setminus (I_0\cup I_1)$ except for having one endpoint in $I_0$ and the other endpoint in $I_1$.

And suppose $A_i\cap A_j=\varnothing$ when $i\neq j$.

We may assume $A_0=[0,1]\times \{0\}$ and $A_1=[0,1]\times \{1\}$.

It seems intuitively true that there should be an arc $A\subseteq [0,1]^2$ between $I_0$ and $I_1$ that misses $\bigcup _{n\in\omega} A_n$.

Is this true?

MY ATTEMPT:

First consider the entire closed connected region $T_0=[0,1]^2$ between $A_0$ and $A_1$ (as defined above).

If there is no other $A_n$ in $T_0$, then we're done (the arc $A$ is easy to find).

Otherwise, choose the least $n>1$ such that $A_n\subseteq T_0$.

Get a tube $T_1$ (homeomorphic to $[0,1]^2$) between $A_1$ and $A_n$. So $T_1\cap A_0=\varnothing$.

Now get $T_2$ by going from $A_n$ to $A_m$, where $m>n$ is least index of any arc $A_m\subseteq T_1$.

We can continue this process to get a nested sequence of compact connected tubes $T_n$, so that $T_n$ misses $A_{n-1}$ for each $n<\omega$.

Then $T:=\bigcap T_n$ is a compact connected set from $I_0$ to $I_1$ that misses all $A_n$'s. But does it contain an arc from $I_0$ to $I_1$?

EDIT (June 23): There are now two answers below, which claim opposite things. I'll try to figure out which argument is flawed, but would appreciate any help. Please note that I proved (above) that there is a continuum between $I_0$ and $I_1$ which misses all arcs $A_n$. So does this mean @Colin 's answer is flawed?

Regarding @Santana 's answer:

Think about a sequence of arcs doing this. So $$A_0=[0,1]\times \{0\}$$ and for $n>0$, $$A_n=\text{graph of the function } f_n:[0,1]\to \mathbb R \text{ where }f_n (x)=\frac{1}{2}|2x-1|+\frac{1}{2^n}.$$

This is not homeomorphic to any collection of horizontal arcs. Moreover this shows you cannot start at an arbitrary remaining endpoint. Every arc beginning at a point in $\{0\}\times (0,1/2)$ and ending in $I_1$ must intersect some $A_n$.

Answer 1

I will argue that the answer is no: there may be no such $A$.

Lemma: there is a countable set of disjoint arcs from $I_0$ to $I_1$ such that any arc from $I_0$ to $I_1$ disjoint from this set has only finitely many $1$'s and $2$'s in the base-4 representation of its startpoint in $I_0$

I will index the arcs using rational numbers $q=a/4^b\in[0,1]$, where $A_q$ will go from $(0,q)$ to $(1,q).$ The arcs will be constructed using induction on $b$ (restricting the numerator $a$ to not be a multiple of $4$, except for $q=0=0/4^0$).

For the base case $b=0$, take a straight line for the arc from $(0,0)$ to $(1,0)$, and a straight line for the arc from $(0,1)$ to $(1,1)$.

Now assume $A_{a/4^b}$ and $A_{(a+1)/4^b}$ are given; these bound an open disc and we are free to choose sufficiently wiggly $A_{(4a+1)/4^{b+1}}$, $A_{(4a+2)/4^{b+1}}$, and $A_{(4a+3)/4^{b+1}}$ in this region to guarantee that the $x$-coordinate of any arc trapped between $A_{(4a+1)/4^{b+1}}$ and $A_{(4a+3)/4^{b+1}}$ switches between $1/4$ and $3/4$ at least $b$ times.

Consider a new arc $A$ with starting point $(0,r)$, disjoint from all the $A_q$ constructed above. By construction, for each $b$, if the $b$'th base-4 digit of $r$ is $1$ or $2$, then $A$ switches between $1/4$ and $3/4$ at least $b$ times. So $r$ cannot have infinitely many $1$'s or $2$'s in its base-4 representation, QED.

Full result

Plonk a scaled copy of the above construction into $[0,1/3]\times [0,1]$, and another copy in $[2/3,1]\times[0,1]$, restricting the second copy to arcs with $q\leq 1/2$. Join the first copy's $A_q$ to the second copy's $A_{q/2}$ by a straight line from $(1/3,q)$ to $(2/3,q/2)$. This rules out all the remaining starting points because the only numbers $q$ such that $q$ and $q/2$ both have only finitely many $1$'s and $2$'s in the base-4 representations are of the form $a/4^b$.

Answer 2

Notation: for any arc $A_n$, we denote endpoints as $e_n:=A_n\cap I_0$ and $v_n:=A_n\cap I_1$.

We consider two separate cases.

Case 1: Suppose that there exist two arcs $A_n$ and $A_m$ such that there is no arc $A_k$ such that $e_n<e_k<e_m$. This implies that $A_n$ and $A_m$ bound an open disk in $[0,1]^2$, and hence there is some arc $A$ that connects $I_0$ and $I_1$ satisfying our conditions.

Case 2: Suppose that for every pair of curves $A_n$ and $A_m$ such that $e_n<e_m$ there must exist another arc $A_k$ such that $e_n<e_k<e_m$. We now consider three reductions of the problem to prove that, in this case, such an arc still exists.

Reduction 1:

We may assume that each $A_n$ is a straight line. If we apply a homotopy that takes each $A_n$ to the straight line between its endpoints, we note that since homotopy classes of arcs on $[0,1]^2$ are identified by endpoints the ordering of the endpoints is preserved. Moreover, this preservation also implies that these straight lines are disjoint. We can imagine the action of this homotopy being similar to taking $I_1$ and slowly moving it to the right, pulling each arc taught.

Reduction 2:

We may assume that the two sets $\{e_n\}_{n\in\mathbb{Z}_+}$ and $\{v_n\}_{n\in\mathbb{Z}_+}$ are each order isomorphic to $(\mathbb{Q}, <)$, and hence that $e_i,v_i\in\mathbb{Q}$.

First, we note that for any two arcs $A_i$ and $A_j$, if $e_i<e_j$ then $v_i<v_j$, otherwise $|A_i\cap A_j|\ne\emptyset$. Moreover, since we are assuming that for any $n,m\in\mathbb{Z}_+$ such that $e_n<e_m$ there exists some $k\in\mathbb{Z}_+$ such that $e_n<e_k<e_m$, this implies that the set $\{e_n\}_{n\in\mathbb{Z}_+}$ is a countable linear order that is order dense. Since we are assuming that $e_i\in(0,1)\times\{0\}$, there is no first or last element. Hence, by Cantor's Theorem, this set is order isomorphic to $(\mathbb{Q},<)$. The same argument applies for the set $\{v_n\}_{n\in\mathbb{Z}_+}$. Applying this isomorphism and composing with the homeomorphism $\mathbb{Q}\to [0,1]\cap\mathbb{Q}$ yields the second claim.

(Note that if we do not assume that $A_0$ and $A_1$ form the top and bottom sides of $[0,1]^2$ respectively, then we can follow these arcs to their endpoints on $I_1$ then find a homeomorphism that takes $[v_0,v_1]$ to $[0,1]$. Hence, this is a safe assumption to make.)

Reduction 3:

We may assume that all arcs are horizontal lines. Using the prior reduction, we shift endpoints -- preserving order -- along both $I_0$ and $I_1$ to form each curve to a horiztonal line.

Final Proof:

Now, the question is this:

Given the arcs $A_x:=\{x\}\times[0,1]$ for every $x\in[0,1]\cap\mathbb{Q}$, is there some arc $A$ with endpoints on $[0,1]\times\{0\}$ and $[0,1]\times\{0\}$ such that $A\cap A_x=\emptyset$ for all $x\in\mathbb{Q}$?

Clearly, the answer is yes! Take any vertical line corresponding to an irrational number $y\in(0,1)$, and this arc will satisfy our conditions.