# The interval $(a,b) \subseteq \mathbb{R}^{2}$ is bounded - metric spaces

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

Bounded:

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.

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

## 1 Answer

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 $$\mathbb{R}$$.