proof

[exotic grail]

How do I write this out

[tame basalt]

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[tired depot]

well, first show that

$\bigcup^{k}_{n=1}(-2n^3,2n^3) = (-2k^3,2k^3)$

then take the limit as k approaches infinity

[silent cliff]

No Lifer #GTFOanemia

[tired depot]

you can do the first bit by showing that

$\bigcup^{m}{n=1}(-2n^3,2n^3) \subset \bigcup^{m+1}{n=1}(-2n^3,2n^3)$

[silent cliff]

No Lifer #GTFOanemia

[tired depot]

induction and all

or just prove that for any n in (-2n^3 , 2n^3) that all values less than n when plugged into (-2n^3 , 2n^3) are a subset of the original

and if A is subset of B

A U B = B

this should be fairly easy to prove

[tired depot]

TP $(-2(n-1)^3,2(n-1)^3) \subset (-2n^3,2n^3) $ where $n \in \mathbb{N}$

we can do this by showing $-2n^3 \leq -2(n-1)^3$ and $2n^3 \geq 2(n-1)^3$ $\forall n\in \mathbb{N}$

$-2n^3 \leq -2(n^3 - 1 - 3n^2 + 3n) $

go on from there

[silent cliff]

No Lifer #GTFOanemia