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Solve for D.

When I tried solving for d I got the following: B1 = { -1 < x < 1 } B2 = { -2 < x < 2 } B3 = { -3 < x < 3 } B4 = { -4 < x < 4 } B5 = { -5 < x < 5 }

I am unsure as to why the answer is { -1 < x < 1 }

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    It is the set of $x$'s which satisfy **all** of the inequalities simultaneously. $\cap$ represents intersection, i.e. $a\in (\bigcap B_i)$ iff $a\in B_i$ for every $i$. – 2017-01-20
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    And for the record, it should be $\{\color{red}{x\in \Bbb Q}~:~ -1 – 2017-01-20
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    The sets $B_i$, $i=1,2,3,4$ you have computed are not correct. Check them! – 2017-01-20
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    @HoneyBee that should be right?? – 2017-01-20
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    As you have written it, it appears as though every real number between $-1$ and $1$, as opposed to only the rational numbers. Also the notation is off – 2017-01-20
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    @JMoravitz i understand the notation being incorrect, if i fix the notation will everything be correct? – 2017-01-20
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    still do not understand as to why the answer is { -1 < x < 1 } – 2017-01-20
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    @Keving if you fix these notations, as pointed out by JMoravitz then the answer is the set $B_1$ – 2017-01-20
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    @HoneyBee But why? – 2017-01-20

1 Answers 1

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$\bigcap_{i=1}^{5}B_i$ is the same as $B_1 \cap B_2 \cap B_3 \cap B_4 \cap B_5$. So in order for an element to be in the intersection of these 5 sets it must be in all of them. It should also be clear that $B_1 \subset B_2 \subset B_3 \subset B_4 \subset B_5$, so the only elements that are in all 5 sets are exactly all of the elements of $B_1$.