Flea and comb are connected $$ X = \{(0,1)\} \cup \{ (x,0) : 0 < x \leq 1\} \cup \left\{\left(\frac{1}{n},y\right) : n \in \mathbb{N} , \; \; 0 \leq y \leq 1\right\} $$
Is $X$ connected? I want to show $X$ is connected but It is kind of hard. Can someone help me solve this problem? 
 A: Let $Y = \{ (x,0) | 0 < x \leq 1\} \cup \{(\frac{1}{n},y) | n \in \mathbb{N} , \; \; 0 \leq y \leq 1\}$. It should be clear that $Y$ is connected since it is path connected (imagine a comb).
You have $X= \{(0,1) \} \cup Y$.
Suppose $X \subset U \cup V$ where $U,V$ are open and disjoint. Without loss of generality, $Y \subset U$ ($Y$ cannot intersect both, otherwise it would not be connected). Suppose $X \subset V$ (that is, $(0,1) \in V$). Since $V$ is open, it must contain an infinite number of the points $\{(\frac{1}{n},1) \}_n \subset Y$, which contradicts $U,V$ being disjoint. Hence $X \subset U$.
Hence $X$ must be connected.
Addendum: For those with masochistic tendencies:
To show that $Y$ is connected without using path connectedness, we use the same technique.
I need a little result first: Suppose $C_1,C_2$ are two connected sets and $C_1 \cap C_2 \neq \emptyset$. Then $C = C_1 \cup C_2$ is connected.
To see this, suppose $U,V$ are disjoint open sets such that $C \subset U \cup V$. 
Let $x \in C_1 \cap C_2$, and without loss of generality, $x \in U$. Since $ x\in C_1$, we must have $C_1 \subset U$ (otherwise this would contradict the connectedness of $C_1$). Similarly we must have $C_2 \subset U$, hence $C \subset U$, and so $C$ must be connected.
Back to $Y$:
I need some notation. Let $T_n = \{ \frac{1}{n} \} \times [0,1]$, $H=(0,1]\times \{0\}$. Clearly $Y = H \cup (\cup_n T_n)$ and $H \cap  T_n \neq \emptyset$ for all $n$.
As above, suppose $U,V$ are disjoint open sets such that $Y \subset U \cup V$. Without loss of generality, suppose $H \subset U$. Now pick one of the $T_n$, then since $T_n \cap H \neq \emptyset$, we must have $T_n \subset U$. Since this is true for all $n$, we have $T_n \subset U$ for all $n$ and hence $Y \subset U$. Hence $Y$ is connected.
