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I have been recently studying topology and it's kind of difficult for me to use the theory in some tasks.I am not sure how to prove that this set is open/closed: $$A=\{(x,y,z) \in \Bbb R^3\mid 0\le x\le 1,\:0\le y\le 1,\:0\le z\le 1\}$$ I know that I have to use a ball and prove something of a kind that for every $x$ in $A$ there must exist an $r>0$ such that the ball with that kind of radius is a subset of $A$ but I am not really sure how to do that.Could someone give me a hand?

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Bring definitions in to play.

To be open, you need an open neighborhood around every point in set. In $R^3$ you can use open balls.

Take $(0,0,0)$ and any ball around it. Is it fully inside the $A$?

If it is $B((0,0,0), \varepsilon)$, then is $(-\frac{\varepsilon}{2},-\frac{\varepsilon}{2},-\frac{\varepsilon}{2}) \in A$?

To be closed, you either look at $A^C$, or you find the limit of every converging sequence, or you find every boundary point and check whether it is in $A$.

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  • $\begingroup$ So If I decide to check every boundary point I need to have a look at all balls with point (0,0,0) , (1,1,1) and etc. is that correct? $\endgroup$ – Thresh Bot Jun 20 '18 at 16:45
  • $\begingroup$ Usually the easiest way to prove a set is closed is to prove its complement is open $\endgroup$ – Sambo Jun 20 '18 at 17:08

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