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

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The union of the infinite sequence of intervals \(\bigcup_{n=1}^{\infty} [5^{-n}, n]\) converges to the interval \((0, \infty)\), as the lower bounds approach 0 without including it, while the upper bounds extend indefinitely. The discussion also explores the intersection \(\bigcap_{n=1}^{\infty} [1+\frac{1}{n}, 1+n]\), which converges to the single point \(\{2\}\). This conclusion is reached by analyzing the bounds as \(n\) approaches infinity, confirming that any \(x\) in the intersection must equal 2. The reasoning highlights the behavior of the bounds in both union and intersection cases. Understanding these concepts is crucial in set theory and real analysis.
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What's the meaning of \displaystyle \bigcup_{n=1}^{\infty} [5^{-n}, n]?
 
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Guest said:
What's the meaning of \displaystyle \bigcup_{n=1}^{\infty} [5^{-n}, n]?

Hi Guest! (Smile)

It's:
$$\bigcup_{n=1}^{\infty} [5^{-n}, n]
=[5^{-1},1] \cup [5^{-2}, 2] \cup [5^{-3}, 3] \cup \dots = (0, \infty)$$
 
I like Serena said:
Hi Guest! Welcome to MHB! (Smile)

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

How did you get $(0, \infty)$?
 
Guest said:
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.
 
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.
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.
 
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.

Yes. The same method applies.
Where are you stuck? (Wondering)
 
I like Serena said:
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.
 
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, $A = \{2\}$.
 
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