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

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Discussion Overview

The discussion revolves around the meaning and implications of the union and intersection of infinite sequences of intervals, specifically focusing on the expressions \(\bigcup_{n=1}^{\infty} [5^{-n}, n]\) and \(\bigcap_{n=1}^{\infty} [1+\frac{1}{n}, 1+n]\). Participants explore the resulting sets and the reasoning behind their conclusions, engaging in both conceptual clarification and mathematical reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • Some participants propose that \(\bigcup_{n=1}^{\infty} [5^{-n}, n] = (0, \infty)\), suggesting that the lower bound approaches 0 but never reaches it, while the upper bound is unbounded.
  • Others express uncertainty about how the conclusion of \((0, \infty)\) is reached, particularly questioning the treatment of the lower bound.
  • One participant suggests applying a similar method to \(\bigcap_{n=1}^{\infty} [1+\frac{1}{n}, 1+n]\) and notes that the answer is \(\{2\}\), but expresses confusion about the reasoning.
  • Another participant provides a detailed explanation of why \(\bigcap_{n=1}^{\infty} [1+\frac{1}{n}, 1+n] = \{2\}\), showing the bounds and confirming that 2 is the only element in the intersection.
  • There is a challenge raised regarding the interpretation of the bounds in the intersection, with one participant mistakenly concluding that it results in \([1, \infty)\).

Areas of Agreement / Disagreement

Participants generally do not reach a consensus on the interpretation of the union of intervals, as some agree with the conclusion of \((0, \infty)\) while others question the reasoning. The intersection discussion appears to have more clarity, with one participant providing a clear argument for \(\{2\}\), but confusion still exists about the method applied.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the behavior of the bounds in both the union and intersection cases. Some participants rely on informal reasoning without fully resolving the mathematical steps involved.

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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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