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

Question

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.

Answer 1

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.

Answer 2

To prove that $\{(0,q):q\in\Bbb Q\cap[0,1]\}\subset U$, note that each point in this set is a limit point of $ T\cup\bigcup_{n}S_n\subset U$, and $V$ cannot contain any limit point of $U$. This completes your proof of connectedness.

Note that $X$ is not even path-connected. Suppose there is a path joining $(0,1/2)$ and $(-1,1)$, i.e., $\exists$ a continuous function $f:[0,1]\to X$ s.t. $f(0)=(0,1/2)$ and $f(1)=(-1,1)$. We use the notation $f(t)=(f_1(t),f_2(t))$. Since $f_1(1)=-1$, we know that for each $x\in (-1,0)$, there exists $t$ s.t. $f_1(t)=x$ by Intermediate Value Theorem. For each $n\in\Bbb N$, we pick $t_n\in (0,1)$ s.t. $f_1(t_n)\in (-\frac{1}{n},-\frac{1}{n+1})$. This forces $f_2(t_n)\in \{1\}\cup (\frac{1}{n+1},\frac{1}{n})$. In any case, we have $t_n\to 0$ as $n\to\infty$, but $f(t_n)=(f_1(t_n),f_2(t_n))\not\to(0,1/2)$, which contradicts the continuity of $f$ at $t=0$.

To see that $Y=\overline X$ is non-contractible, consider the map $r$ that projects everything with negative $x$-coordinate to the $y$-axis and fixes everything in the first quadrant or on the axes. This is a retraction onto the triangle having $(0,0)$, $(1,1)$ and $(0,1)$ as vertices. Since the triangle is homeomorphic to $S^1$, we have a surjective homomorphism $r_\ast:\pi_1(Y)\to\pi_1(S^1)$ which suggests that $\pi_1(Y)$ is non-trivial. Hence, $Y$ is non-contractible. (Note that $Y$ is actually path-connected so we don't have to specify a base point.)

Answer 3

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 $X$.