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My task is this:
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!
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.
