What does the union of an infinite sequence of intervals converge to?

[Guest2]

What's the meaning of $$\displaystyle \bigcup_{n=1}^{\infty} [5^{-n}, n]$$?

 

[I like Serena]

What's the meaning of $$\displaystyle \bigcup_{n=1}^{\infty} [5^{-n}, n]$$?

Hi Guest! (Smile)

It's:
$$\bigcup_{n=1}^{\infty} [5^{-n}, n]
=[5^{-1},1] \cup [5^{-2}, 2] \cup [5^{-3}, 3] \cup \dots = (0, \infty)$$

 

[Guest2]

Hi Guest! Welcome to MHB! (Smile)

It's:
$$\bigcup_{n=1}^{\infty} [5^{-n}, n]
=[5^{-1},1] \cup [5^{-2}, 2] \cup [5^{-3}, 3] \cup \dots = (0, \infty)$$

Hi, thanks for replying.

How did you get $(0, \infty)$?

 

[I like Serena]

Hi, thanks for replying.

How did you get $(0, \infty)$?

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.

 

[Guest2]

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.

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.

 

[I like Serena]

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.

Yes. The same method applies.
Where are you stuck? (Wondering)

 

[Guest2]

Yes. The same method applies.
Where are you stuck? (Wondering)

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.

 

[Euge]

Hi Guest,

Here's a way to show that

$$\bigcap_{n = 1}^\infty \left[1 + \frac{1}{n},1 + n\right] = \{2\}.$$

Let $A$ represent the set on the left hand side of the above equation. If $x\in A$, then

$$1 + \frac{1}{n} \le x \le 1 + n\quad \text{for all} \quad n\in \Bbb N.$$

In particular, setting $n = 1$, we get

$$2 \le x \le 2.$$

This means that $x = 2$. Hence $A \subseteq \{2\}$. On the other hand, we know that if $n\in \Bbb N$, $\frac{1}{n} \le 1 \le n$, which implies

$$1 + \frac{1}{n} \le 2 \le 1 + n.$$

Since $n$ was an arbitrary natural number, then $2 \in A$. Therefore, $A = \{2\}$.