$C = (\{0\} \times [0,1]) \: \cup \: \{ (\frac{1}{n},y), n \in \mathbb{N}, y \in [0,1] \} \:\cup \: ([0,1] \times \{0\})$ is connected?

Question

I want to prove wich $C = (\{0\} \times [0,1]) \: \cup \: \{ (\frac{1}{n},y), n \in \mathbb{N}, y \in [0,1] \} \:\cup \: ([0,1] \times \{0\})$ is connected.

My attempt:

$C$ is the union of sets. If we show that the sets have non-empty intercept, then $C$ is connected.

Let $A = (\{0\} \times [0,1])$ and $B = \{ (\frac{1}{n},y), n \in \mathbb{N}, y \in [0,1] \} \:\cup \: ([0,1] \times \{0\})$. $A \cap B = \{(0,0)\}$

$A$ is connected? Let $f: [0,1] \rightarrow \mathbb{R}^2$ such that $f(x) = (0,x)$, $f$ f is a continuous function from point $(0,0)$ to point $(0,1)$. Then, it is a path with a starting point and an ending point. Therefore, $A$ connected by paths, this implies $A$ connected.

$B$ is connected? Let $S= \{ (\frac{1}{n},y), n \in \mathbb{N}, y \in [0,1] \}$ and $T = ([0,1] \times \{0\})$. The intercept are points of the type $ (1/y,0), y \in \mathbb{N}$. T is connected, in an analogous way we draw from $A$. For the set $S$, we can create a function for each vertical line with the joining points $(1/y,0)$ and from there conclude that it is path connected. Therefore, $B$ is connected.

This is enough to show that $C$ is connected.

I feel that the justification of $B$ being related is not quite right, but I don't understand why.

I really appreciate any help.

Answer 1

You can use that the union of two connected subsets of nonempty intersection is connected.

So $A\cup T$ is connected.

For $k\in \mathbb{N}^{\ast}$, let $S_k=\{\left(\frac1k, y\right)|y\in[0,1]\}$

We have $C=A\cup T\cup \bigcup_{k=1}^{+\infty} S_k$

By induction, we can prove that forall n, $C_n = A\cup T\cup \bigcup_{k=1}^{n} S_k$ is connected.

Finally, let X and Y two points of C, there exists $N\in\mathbb{N}^{\ast}$ such that X and Y are in $C_N$. However $C_N$ is connected and we can find a path in $C_N$ then in $C$ from X to Y.

Answer 2

I think that it is much easier to prove directly that $C$ is path connected. First of all, if $(x,y)\in C$, then $(x,t)\in C$ for each $t\in[0,y]$, and therefore there is a path in $C$ going from $(x,y)$ to $(x,0)$. And of course, if $(x,0),(y,0)\in C$, then $(t,0)\in C$ for each $t$ between $x$ and $y$, and therefore there is a path in $C$ going from $(x,0)$ to $(y,0)$. Finally, if $(x_1,y_1),(x_2,y_2)\in C$, then you first go in a straight line from $(x_1,y_1)$ to $(x_1,0)$, then from $(x_1,0)$ to $(x_2,0)$ and finally from $(x_2,0)$ to $(x_2,y_2)$.

Answer 3

Let $f:C \to \{0,1\}$ be a continuous function. Since $[0,1]\times \{0\}$ is connected, $f|_{[0,1]\times \{0\}}$ is constant (say equal to $c$). Then, for $x\in \{1/n: n\in\mathbb N\} \cup \{0\}$, as $\{x\}\times [0,1]$ is connected, we know that $f$ is constant on it equal to $f(x,0) = c$.

Therefore $f$ is constant on $C$. This being true for any function $f$, we see that $C$ is connected.