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Given an open subset $I \subset \mathbb{R}$ if $(x_1-r_1,x_1+r_1) \subset I$ does that imply that $[x_1-r_1,x_1+r_1] \subset I$? If not, could I choose $r_1>0$ that such that this holds?

More generally, Let $X$ be a complete metric space. If $W$ is open and $B(x_1,r_1) \subset W$, does it imply that $B[x_1,r_1] \subset W$? Could I choose such a $r_1$?


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If $I=(0,1)$ what happens? – Ross Millikan Jan 20 '12 at 5:25
up vote 5 down vote accepted

Your first question: no. Take $I=(-1,1)$, $x_1=0$, $r_1=1$. Then certainly $(-1,1)\subseteq I$, but it is false that $[-1,1]\subseteq I$.

However, if $(x_1-r_1,x_1+r_1)\subset I$ and $r_1\gt 0$, can you show that if you replace $r_1$ with, say, $\frac{r_1}{2}$, then it will work?

(You don't even need $I$ to be open: just note that $$\left\{y\;\left|\; d(x_1,y)\leq \frac{r_1}{2}\right\}\right. \subseteq \{y\mid d(x_1,y)\lt r_1\}$$ and this holds for any metric space, whether complete or not, when $r_1\gt 0$).

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Absolutely, Thanks for your Answer. – Jr. Jan 20 '12 at 5:42

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