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I have a question. I have to check whether $0 \in \Bbb R^2$ is an interior point or a boundary point of the following set $$W = \Bbb R^2_{++} = \{(x_1,x_2) \in \Bbb R^2 \mid x_1 > 0, x_2 > 0\}.$$

I thought maybe it is a boundary point, because any open disc with center $(0,0)$ contains at least one point of $W$ and at least one point of the complement of $W$. Is this correct? And how can I prove it, or is this reasonable enough?

Thank you

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    Usually if you say "any open disc with center (0,0) contains at least one point of W and at least one point of the complement of W" it is enough, but formally you should be able to give a proof of this statement, like saying that $(a,a)\in W$ in your open disc for small enough positive $a$ and $(a,a)\notin W$ for small enough negative $a$. – 2017-01-30
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    Yes, your argument is fine. Just some little suggestions: 1. I'd rather write $(0,0) \in \Bbb R^2$. 2. You may use `$\Bbb R^2 \setminus W$` to show $\Bbb R^2 \setminus W$. – 2017-01-30

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Of course it cannot be an interior point since it is not a point of the set. It is a boundary point and your proof is correct. To make it perfectly formal you should say, given a radius $R>0$, which is the point of the set which has distance less than $R$ from $0$.