Show that $Y$ is not path-connected

Question

Let $\mathbb{R}^2$ with the usual topology and let

$$ Y = A_0 \cup (\bigcup_{n \in \mathbb{N}} A_n) \cup (\bigcup_{n \in \mathbb{N}}L_n)$$

where

$$ A_0 = \{ 0 \} \times [0,1] \qquad A_n = \{ {\dfrac{1}{n}} \} \times [0,1] \qquad \textrm{and $L_n$ is the segment between $(\dfrac{1}{n},1)$ and $(\dfrac{1}{n+1},0)$}$$

I need to show that $Y$ is not path-connected.

My intuition is that there cannot be any continuous function $f:[0,1] \to Y$ such that (for example) $f(0) = (0,0)$ and $f(1) = (1,1)$ because it is not going to be continuous at $(0,0)$, but I cannot prove that formally.

Answer 1

Your intuition is good. To make it precise, let $\pi_1,\pi_2$ denote the coordinate projections $\mathbb{R}^2 \to \mathbb{R}$. Then suppose there were a continuous $f\colon [0,1]\to Y$ with $f(0) = (0,0)$ and $f(1) = (1,1)$. Let $g = \pi_1\circ f$, and define

$$t_n = \inf \left\{ t \in [0,1] : g(t) = \frac{1}{n}\right\}.$$

Looking at $(f(t_n))$ leads to a contradiction.