I am trying to show that the interval $(a,b) \subseteq \mathbb{R}^{2}$ is a bounded set.

By $(a,b) \subseteq \mathbb{R}^{2}$ I am meaning $(a,b) \times \{0\} = \{(x,y) \in \mathbb{R}^{2}: a<x<b, y=0\}$


If I can show that $(a,b) \times \{0\}$ is contained in some open/closed ball then I am done. Would the following work? $B((\frac{a+b}{2}, 0), 5(b-a))$ i.e. a ball centred at $(\frac{a+b}{2}, 0)$ with a radius 5 times the length of the interval.

I'm getting a bit confused because we are in $\mathbb{R}^{2}$.

Also, am I correct in thinking that the interval $(a,b)$ is not open as a subset of $\mathbb{R}^{2}$ but it is open as a subset of $\mathbb{R}$? At least that is what I seem to have proven.

  • $\begingroup$ This all looks correct to me $\endgroup$ – Greg Martin Jun 20 '20 at 17:38
  • $\begingroup$ Everything looks good to me. (But $5(b-a)$ is overkill $-$ $\frac12(b-a)$ is enough.) $\endgroup$ – TonyK Jun 20 '20 at 17:39
  • $\begingroup$ Thanks both. Yes I was over enthusiastic in my selection... $\endgroup$ – Mathlearner Jun 20 '20 at 17:40
  • $\begingroup$ @gen-zreadytoperish: The OP already explained it, in the second line of the question. $\endgroup$ – TonyK Jun 22 '20 at 13:50

Yes, your ball will work. You could also have taken the ball centered at the origin with radius $b$, i.e. $B((0,0),b)$.

You are correct that $(a,b)$ is not open in $\mathbb{R}^2$, but is open in $\mathbb{R}$.


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