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?

2 Answers

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 $$|y_n-x|<dfrac{r}{2}$$ for every $n$. Can you go to the limit as $nrightarrowinfty$ and conclude?

Answered by Olivier Moschetta on November 29, 2020

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