I want to show that $I = \{x \in \mathbb{R}^3 : 1 \leq x_1 \leq 3, 0 \leq x_2, x_3 \leq -1\}$ is closed by showing that the complement is open with open balls.
But how can I do that? Thanks
Show that a set is closed 2
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general-topology
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0I have done this: you cannot say that there is for any point x in I an open ball with center z and radius r. Because if we take X1=3, then we cannot find a r such that ||x-z|| < r. But I want to show that it is closed, so I thought that R3\I is open. But I don't know how to do that. – 2017-01-30
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0Can you write down which elements are in $\mathbb{R}^3 \setminus I$? – 2017-01-30
1 Answers
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If you want to show, that a subset $A$ of some metric space is open, you pick a point $x\in A$ and try to find such radius $r_x>0$, that an open ball $B(x, r_x)$ is contained in $A$. It's a basic procedure in all metric spaces and here it will work exactly the same way.
You need to take an arbitrary point $x\in (\mathbb{R}^3 \setminus I)$ and show that for some $r>0$ a set of form $\{y\in \mathbb{R}^3: \|x-y\|
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0But don't you have to check it for every x or is it enough to do it for one? And how can you do it for every x? – 2017-01-30
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0You do have to check it for every $x$. However, notice that taking the distance between our $x\in (\mathbb{R}^3 \setminus I)$ and $I$ suffices, since the inequality in open ball (the set I've mentioned earlier) is strict. Or you can take half of this distance, just to make sure. – 2017-02-09