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I want to prove that an interior point of $B=\{(x_1,x_2), x_1> 0, x_2> 0\} \subset \mathbb{R}$ is in the set

$A=\{(x_1,x_2), x_1\ge 0, x_2\ge 0\}\subset \mathbb{R}$

For this I've picked a generic point from the set $B$ and imagined an open disk arount it: $D((x_1,x_2),R)$

where $R$ is the radius of the open disk.

I need to prove that every point of this open disk is in $B$, so I've imagined a point $y=(y_1,y_2)$ and I'm trying to prove that $$|y-x|<R$$ I guess.

I need help for this, and I'd like someone to help me.

Thanks.

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3 Answers 3

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Hint: You want to pick an $R$ for each $x$ such that the disk centered at $x$ of radius $R$ is inside $B$. (Any arbitrary $R$ will not work - you have to pick it.) To figure out what your $R$ must be, consider how close your point $x$ is to each axis.

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    $\begingroup$ I've figured that $R$ must be $min(x_1,x_2)$ but I can't go any further $\endgroup$ May 8, 2015 at 3:37
  • $\begingroup$ Excellent! You know that your disk is actually going to live inside of an open square with side lengths $2R$ centered at $x$. (Draw it if you can't tell!) What can you say about this square? $\endgroup$ May 8, 2015 at 3:39
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You don't need to prove that $|y-x| < R$. Rather, you pick $y$ such that $|y-x| < R$. You can then show that $y \in B$, but this depends on the choice of $R$.

So, from a good initial choice of $R$, you can show that if $y$ is in the open disk of radius $R$ around $x$, then $y \in B$. You can finally show that $y \in A$.

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  • $\begingroup$ shouldn't I need to show that $y\in A$? $\endgroup$ May 8, 2015 at 3:40
  • $\begingroup$ @GuerlandoOCs Yes, you eventually need to show that, as I wrote in my last sentence. As for first showing that $y \in B$, this is a reasonable first step. You wrote yourself: "I need to prove that every point of this open disk is in B." $\endgroup$
    – Théophile
    May 8, 2015 at 4:22
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An alternative way is to observe that $B$ is open and contained in $A$. Hence $\operatorname{Int}(B)\subset B\subset A$.

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