# How to show closure of ball of radius r/2 is a subset of ball with radius r

Mathematics Asked by hi im epsilon on November 29, 2020

I’m trying to show that the neighborhood $$overline{N_{r/2}(x)}$$ (the bar is really small but that’s the closure) is a subset of $$N_{r}(x)$$.

I split it up so that I know if $$y in overline{N_{r/2}(x)}$$ then y is either a limit point or an element of $$N_{r/2}(x)$$. If it’s an element then it’s pretty straightforward but I don’t know what to do if it’s a limit point. Maybe I’m doing this the wrong way. Any ideas?

This echoes the other answer but, the closure of the open $$r/2$$ ball centered at $$0$$, is pretty easily nothing but the closed $$r/2$$ ball. Clearly the latter is contained in the $$r$$ ball. That is, if we're in $$Bbb R^n$$.

Answered by Chris Custer on November 29, 2020

I'm interpreting $$N_{r}(x)$$ as the ball centered at $$x$$ of radius $$r$$ in a metric space. Correct me if I'm wrong there.

If $$yinoverline{N_{r/2}(x)}$$ then there exists a sequence $$y_n$$ such that $$y_nin N_{r/2}(x)text{ and } y_nrightarrow y$$ We therefore have $$|y_n-x| for every $$n$$. Can you go to the limit as $$nrightarrowinfty$$ and conclude?

Answered by Olivier Moschetta on November 29, 2020