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What's the meaning of \displaystyle \bigcup_{n=1}^{\infty} [5^{-n}, n]?
Guest said:What's the meaning of \displaystyle \bigcup_{n=1}^{\infty} [5^{-n}, n]?
Hi, thanks for replying.I like Serena said:Hi Guest! Welcome to MHB! (Smile)
It's:
$$\bigcup_{n=1}^{\infty} [5^{-n}, n]
=[5^{-1},1] \cup [5^{-2}, 2] \cup [5^{-3}, 3] \cup \dots = (0, \infty)$$
Guest said:Hi, thanks for replying.
How did you get $(0, \infty)$?
Thank you.I like Serena said:Were getting an interval with a lower bound that is the result of $5^{-1},5^{-2},5^{-3},...$.
It never quite reaches $0$, but it comes closer than any positive value.
Therefore the lower bound of the interval is $0$ while excluding $0$ itself.
The same holds for the upper bound.
Guest said:Thank you.
Could we apply the same method to $\bigcap_{n=1}^{\infty} [1+\frac{1}{n}, 1+n]
$? The answer is $\left\{2\right\}$, but I don't know how to get it.
As n goes to infinity the lower bound $1+1/n \to 1$ and the upper bound $n+1 \to \infty$ so I get $[1, \infty)$ which is wrong.I like Serena said:Yes. The same method applies.
Where are you stuck? (Wondering)