# Prove that the boundary of a square is connected

Show that $\partial([0,1]\times[0,1] \subset \Bbb R^2)$ is connected

My thought is that I can use the theorem that continuous image of a connected set is connected - thus I can build a continuous function that maps an interval [a,b] to $\partial([0,1]\times[0,1] \subset \Bbb R^2)$, and prove the boudary is connected. But how can such a funtion be possible and how to prove it is continuous? Also any other way to show the connectedness is welcomed. Thank you!

• It might be easier to show it is path connected. Jul 8, 2017 at 19:06
• @SahibaArora that makes sense! But path-connected is also defined based on existence of continuous function isn't it? Can you please give more clues on how to show path connectedness? Jul 8, 2017 at 19:13
• Let $f(x)=(x,0)$ for $x\in [0,1].$ Let $f(x)=(1,x-1)$ for $x\in [1,2].$ Let $f(x)=(3-x,1)$ for $x\in [2,3].$ Let $f(x)=(0, 4-x)$ for $x\in [3,4].$ Then $f:[0,4]\to \partial ([0,1]^2)$ is a continuous surjection, demonstrating connectedness. And if $0\leq x_1<x_2\leq 4$ then $f|_{[x_1,x_2]}$ is a path from $f(x_1)$ to $f(x_2).$ Jul 9, 2017 at 2:46
• @DanielWainfleet this is super helpful! Thank you. Jul 9, 2017 at 21:51

Let $f_1:[0,1]\to [0,1]\times[0,1]$ be given by $f_1(x)=(x,0)$. Then $f_1$ is continuous and the image of $f_1$ is connected because $[0,1]$ is connected, i.e. $A_1=\{(x,0):x\in[0,1]\}$ is connected.
Next let $f_2:[0,1]\to [0,1]\times [0,1]$ be given by $f_2(x)=(1,x)$. Then $f_2$ is continuous and the image of $f_2$ is connected because $[0,1]$ is connected, i.e. $A_2=\{(1,x):x\in [0,1]\}$ is connected.
This way we can show that $A_3=\{(x,1): x\in[0,1]\}$, $A_4=\{(0,x):x \in[0,1]\}$ are connected.
Next we observe that $B=\cup_i A_i$ and $A_1\cap A_2,\ A_2\cap A_3, A_3\cap A_4, A_4\cap A_1\neq \emptyset$, where $B=\partial([0,1]\times[0,1])$ and we are done.
$$f:[0,2\pi)\to\partial\left([-1,1]\times[-1,1]\right),\qquad f(\theta)=\frac{(\cos\theta,\sin\theta)}{\max\left(|\cos\theta|,|\sin\theta|\right)}$$ is a continuous map, hence $\partial\left([-1,1]\times[-1,1]\right)=f([0,2\pi))$ is (arc-)connected because $[0,2\pi)$ is (arc-)connected.