Properties of space $X = T \cup \bigcup_{n=1}^{\infty} S_n \cup \{ (0,q) : q \in \mathbb Q \wedge q \in [0,1] \} $

Let $S_n$ be closed interval in euclidean plane $\mathbb R^2$ with ends in $(\frac{-1}{n}, \frac{1}{n})$ and $(\frac{-1}{n}, 1)$.
Moreover let $T$ be boundary of triangle with vertices $(1,1),(0,0),(-1,1)$.
A. Prove that subspace of plane $$ X = T \cup \bigcup_{n=1}^{\infty} S_n \cup \{ (0,q) : q \in \mathbb Q \wedge q \in [0,1] \} $$ is connected space but not arc connectedness
B. Let $Y = \bar{X}$ - closure of the set $X$ in euclidean plane. Prove that $Y$ is not contractible space.

My attempt:

A. Let $U,V$ be open sets such that $U,V \subset \mathbb R^2$ and $X \subset U \cup V$. Assume that $(U \cap X) \cap (V \cap X)= \emptyset$. Now we want to prove that $$(U \cap X) = \emptyset \mbox{ or } (V \cap X)= \emptyset $$

If $ X \subset U \cup V$ then $T \subset U \cup V$. So $T \cap V \neq \emptyset $ or $T \cap U \neq \emptyset$. Without lost of generality assume that $T\cap U \neq \emptyset$. Then $T \cap V = \emptyset$ because $T$ jest connected ($T$ is sum of three intervals and they create triangle and interval is connected). So $T \subset U $.

And that's why $\forall_{n} S_n \cap U \neq \emptyset $ because boundarie of triangle $T$ are ends of intervals from $\bigcup _{n=1}^{\infty} S_n$. Each interval is connected. Let $(\frac{-1}{n}, 1) \in \bigcup _{n=1}^{\infty} S_n$ and $J= I ((\frac{-1}{n}, \frac{1}{n}), (\frac{-1}{n}, 1))$. We know that $J \subset U \cup V$ and $J \cap U \neq \emptyset$ . That's why $J \cap V = \emptyset$ because $J$ is connected, so $J \subset U $. So we have that $T \cup \bigcup_{n=1}^{\infty} S_n \subset U$.

My problems:

Firstly, I want somebody to check my solution.

Secondly, I am not sure how can I prove that:

$$ \{ (0,q) | q \in \mathbb Q \wedge q \in [0,1] \} \subset U$$

If it comes to B. I didn't find any idea how to solve that.

Solution 1:

Notation: $<x,y>$ denotes an ordered pair. Other brackets do not.

Exercise: In any space, if $F$ is a family of connected subspaces and $\cap F$ is not empty then $\cup F$ is connected.

So each $T\cup S_n=\cup \{T,S_n\}$ is connected because $T$ and $S_n$ are connected and $T\cap S_n$ is not empty.

So $T\cup (\cup_{n\in \Bbb N}S_n)=\cup \{T\cup S_n: n\in \Bbb N\}$ is connected because $\cap \{T\cup S_n: n\in \Bbb N\}=T$ is not empty.

For brevity let $W=T\cup (\cup_{n\in \Bbb N}S_n).$

Let $A,B$ be disjoint open subsets of $X$ with $A\cup B=X.$ Since $W$ is a connected subspace of $X,$ we have $W\subset A$ or $W\subset B$.... WLOG let $W\subset A.$

We have $<0,0>\in T\subset A.$ We now show that $\{0\}\times [(0,1]\cap \Bbb Q]\subset A.$

Suppose by contradiction that $q\in \Bbb Q\cap (0,1]$ and $<0,q>\in B.$ There must exist $r\in \Bbb R$ with $0<r<q/2$ such that $B\supset X\cap [(-r,r)\times (q-r,q+r)].$ But if $n\in \Bbb N$ is large enough that $1/n<r$ then, since we also have $1/n<q-r,$ we obtain $$B\cap A\supseteq B\cap S_n\supseteq$$ $$\supseteq [(-r,r)\times (q-r,q+r)]\cap S_n\supseteq$$ $$ \supseteq\{-1/n\}\times (q-r,q)\ne \emptyset$$ which is absurd.

Therefore $X$ is connected.

Addendum. Exercise: A space with a dense connected subspace is a connected space..... Now in the Q, $W$ is a dense connected subspace of $X$.

Solution 2:

Let r,s,t be three irrationals with $1/2 < r < s < 1$, $0 < t < 1/10$.
$X \cap (-t,t)×(r,s)$ = X $\cap$ $[-t,t] \times [r,s]$ is a clopen subset of $X$.
$X$ is not connected.

Solution 3:

A. Your proof is ok but for the last set we cannot prove it individually.

Let $Q=\{(0,q)|q\in\mathbb{Q}\cap[0,1]\}$, and take advantages of the hausdorff properties of $\mathbb{R}^2$.

Consider a horizontal sequence that converges to $(0,q),q\in\mathbb{Q}\cap[0,1]$ i.e. $(x_n)_{n\in\mathbb{N}}=(\frac{-1}{n},q)\to(0,q)$, every point in $Q$ is a limit point of $\bigcup_{n=1}^\infty S_n$, so $\bigcup_{n=1}^\infty S_n\cup Q$ is connected. If not, then $\exists$ open sets $M,N\subset\mathbb{R}^2$ s.t. $M,N$ are disjoint and each component intersect our set. but since $Q$ contains the limit points so if $\bigcup_{n=1}^\infty S_n\subset M$ then $Q\subset M$, they cannot form a separation, contrary to our assumption.

Therefore, $X$ is connected because the two part are both connected and have some points in common.

Thanks @DanielWainfleet for pointing out my flaw in the previous proof.

For the non-arcwise-connectedness, it suffices to show that a path $f:[0,1]\to X$ is not a homeomorphism. (i.e. the path is not homeomorphic to $[0,1]$) Choose an end point $p\in T\cup\bigcup_{n=1}^\infty S_n$ and another end point $q\in Q-\{(0,0),(0,1)\}$. To go to $(0,q)$, we need to path infinitely many vertical segments, just like the topologist's sine curve.

Let $f(t)=(x(t),y(t))$ and to fix idea $x(0)=0$ and $(x(1),y(1))=p$. We need to find a sequence $(t_n)\to 0$ s.t. $y(t_n)\not\to q$.

Given a large $n$ and choose a $x(\frac{1}{n})<u<0$ s.t. $y(u)=1\text{ or }\frac{1}{n}$. By the assumption, $f$ is continuous, so it has the intermediate value property which ensures that the sequence $0<t_n<\frac{1}{n}$ s.t. $x(t_n)=u$. Now we reach a contradiction because there doesn't exist such a continuous path while we assumed that $f$ is continuous. Thus, $X$ cannot be arc-connected.

B.

$Y=X\cup \{(0,q)|q\in[0,1]\subset\mathbb{R}\}$ since $Q$ is dense in the closed unit interval.

And, you can regard $Y$ as infinitely many circles made by ropes that each two of them have at least one point in common $(\bigcup_i S_i^1)$ i.e. $Y\simeq\bigcup_{i\in N} S_i^1$ s.t. $S_j^1\cap S_{j+1}^1\neq\varnothing$ for any $j\in N$.

If $Y$ is contractible then $\pi_t:Y\times [0,1]\to Y$ is nulhomotopic, i.e. homotopy equivalent to a point. So let's assume that $Y$ is contractible.

Choose an arbitrary point $x_0\in Y$ to be the range and $x_0\in S_j^1$, then if we want some point $y$ maps to $x_0$ through $\pi_t$ then there are only two directions to choose, so we choose one. Then, by our assumption, all points except $x_0$ need to move in the same direction. If not, then at some point $k\in Y$, the mapping will be discontinuous since the point on left side will go to the opposite direction of those points on the right hand side of $k$ so a sequence converging to $k$ under $\pi(x,0)$ but will not when $t\in(0,1)$.

To fix idea, let's assume that all point will be mapped continuously to $x_0$ from the right hand side of $x_0$ by $\pi_t$. Now, we have a problem, this rope will be broken near $x_0$ because points on the left hand side will move away while $x_0$ is fixed. So this continuous map $\pi_t$ doesn't exist.

Therefore $Y$ is not contractible.