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I have a question. I have to prove that the following set is open $$I = \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_1 < 1 \vee x_1 > 3 \vee x_2 < 0 \vee x_3 > -1\}$$ But I don't know how to do it.

I think you can check it with open balls, but how can you do it for every $x$? Or can you do it with an epsilon definition?

Thank you

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    But how can you prove it for every x? You can calculate a few x but you cannot do it for every x. How must you do that? – 2017-01-31
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    It is probably with "," – 2017-01-31
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    Your inequalities on $x_{1}$ are asking for $x_{1}<1$ and $x_{1}>3$, which cannot be. Hence your set is empty, which is open. If your inequalities should be in a different way, you need to work with open balls. Suppose the inequalities are $x_{1}\in(1,3)$. Then pick $(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}$ such that $x_{1}\in(1,3)$, $x_{2}<0$ and $x_{3}>-1$ and find open ball $B(x,\epsilon)$ such that $B(x,\epsilon)\subset I$. The fact that you can make $\epsilon$ arbitrarily small will help. – 2017-01-31
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    In the way it is formulated, your question seems very similar to the ones asked by user410573. I strongly suspect you are the same person, and you are trying to make us solve your homework. Please learn how to use this site properly – 2017-01-31

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Your set can be described as $$ I=I_1\cup I_2\cup I_3\cup I_4 $$ where \begin{align} I_1 &= \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_1 < 1\}\\ I_2 &= \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_1 > 3\}\\ I_3 &= \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_2 < 0\}\\ I_4 &= \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_3 > -1\} \end{align}

Now you want to see that each of those sets is open; once done, you get the conclusion, because a union of open sets is open.

Suppose $x=(x_1,x_2,x_3)\in I_1$; then $x_1<1$. Prove you can find $r>0$ such that $$ B(x,r)\subset I_1 $$

Can you do similarly for the other sets?

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    Thank you I almost get it, but for example by I2 have I choose r = (x1-3)/2. – 2017-01-31
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    @C.... Yes; or $(x_1-3)/42$; there's no magic in $2$, just use a value that ensures the ball won't meet the “boundary”. – 2017-01-31
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    I understand about the 2, but if I fill in (X1-3)/2, I get x < 2x1 -3 I think there is going something wrong – 2017-01-31
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    @C.... Maybe you have to adjust the factor, I didn't want to do all computations. Use the fact that the map $(x_1,x_2,x_3)\mapsto x_1$ is continuous. – 2017-01-31
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    I have to do it with absolute values. But by I3, you can't say |x2|<0, how must you do it here? Or is there another method? – 2017-01-31