# Connected and not path-connected

We have $\mathbb R^2$ (real plane) with the Euclidean topology.

Define $X(n) = \{1/n\} \times [-n,n]$, all subspaces of $\mathbb R^2$.

$$Y = \mathbb R^2 \setminus \bigcup_{n\ge1}X(n)$$

Prove that $Y$ is connected but not path-connected. How can I prove this?

Thanks in advance!

• – Grigory M Jan 5 '14 at 15:34

## 2 Answers

Hints.

For path-connectedness: Any path is the continuous image of the compact set $[0,1]$, hence bounded.

For connectedness: Show that a continuous function $f \colon Y \to \{0,1\}$ must be constant.

Proof that $Y$ is not path-connected:

Consider $(x_0,y_0) = (0,0)$ and $(x_1,y_1) = (2,0)$. These are points in $Y$. Let $\gamma(t)$ be a path in $\mathbb{R}^2$ from $(x_0,y_0)$ to $(x_1,y_1)$ ($t$ from $0$ to $1$). Write $\gamma(t) = (x(t),y(t))$ and remark that $y(t)$ is bounded in absolute value, say by an integer $N$. Then $x(t)$ goes from $0$ to $2$. Hence, there exists a $t_0$ such that $x(t_0) = 1/N$. This shows that $\gamma(t_0)$ is not in $Y$.

Proof that $Y$ is connected:

Write $Y$ as $A \cup B$ where the union is disjoint and $A = \{(x,y) \in Y: x \leq 0\}$ and $B = Y - A$. It is easy to show that $A$ and $B$ are path-connected, hence connected. By general topology, the adherence $\bar{B}$ of $B$ in $Y$ is also connected. Hence $A$ and $\bar{B}$ are both connected subspaces which intersect, non-trivially (in the vertical axis). This shows that $Y = A \cup \bar{B}$ is connected.

• This would be a very good answer, if the problem in the question was not of homework type or had the OP shown some effort of their own. – Carsten S Jan 5 '14 at 15:35
• This isn't a homework. This was a question I had on my exam: Analytical Topology and was interested wether my answer was correct or not. – user84112 Jan 5 '14 at 15:36
• @user84112: If your real question is "[Was] my answer correct or not?", then this almost surely fails to be an answer to the question. But it is a [partial] answer to the question you posted, so I have a hard time faulting Jeremy. (Also, I have no qualms with saying most exam problems, including this one, are "of homework type", though the rest of the community may not agree on this point.) – Eric Stucky Jan 5 '14 at 19:18