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What is $\bigcup_{n=1}^\infty[n,n+1]$? What is $\bigcup_{n=1}^\infty(n,n+2)$?

What is $\bigcup_{n=1}^\infty(n,n+1)$? What is $\bigcup_{n=1}^\infty(1/n,1]$?

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I don't really get why the brackets matter. I know [] is inclusive and () is not inclusive but does it make a difference?

B) For the first one, the sets are $[1,2], [2,3], [3,4],\ldots$ so is the set just $[1,2,\ldots\infty)$?

For the second one is it just $(1,2,3,\ldots\infty)$?

C) same thing as the b? but $(1, 2, \ldots, \infty)$

$(0, \ldots , 1/3, 1/2, 1]$

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Your notations $[1,2,\ldots\infty)$ and $(1,2,\ldots\infty)$ don’t really make sense: the interval notation has only two entries, the left endpoint of the interval and the right endpoint. If for the first question you’re asking whether

$$\bigcup_{n=1}^\infty[n,n+1]=[1,\infty)\;,$$

the set of all real numbers $x$ satisfying the inequality $x\ge 1$, the answer is yes. And if for the second part of that question you’re asking whether

$$\bigcup_{n=1}^\infty(n,n+2)=(1,\infty)\;,$$

the set of all real numbers strictly greater than $1$, that answer is also yes, thanks to the fact that successive intervals overlap.

$\bigcup_{n=1}^\infty(n,n+1)$ is another story, however. It may be helpful to start writing it out longhand, so to speak: it’s

$$(1,2)\cup(2,3)\cup(3,4)\cup(4,5)\cup\ldots\;.$$

Recall that an interval $(a,b)$ does not include $a$ or $b$. Thus, neither $(1,2)$ nor $(2,3)$ contains the number $2$, and certainly $2$ is not in $(n,n+1)$ when $n>3$. Similarly, neither $(2,3)$ nor $(3,4)$ contains $3$, and it’s clear that none of the other intervals does, either. In fact, this union doesn’t contain any integers: it just contains the numbers lying strictly between consecutive positive integers. You could write it $(1,\infty)\setminus\Bbb Z^+$, where $(1,\infty)$ is the set of real numbers larger than $1$, and $\Bbb Z^+$ is the set of positive integers. (In case you’ve not seen the notation, $A\setminus B$ means the set of things that are in $A$ but not in $B$. Elementary texts sometimes write it $A-B$ instead.)

Finally, you may have the right idea in the last part, but if so, you’ve written it incorrectly: the correct answer is $(0,1]$. It’s pretty clear that the union doesn’t contain $0$, any negative number, or any number larger than $1$. On the other hand, if $0<x\le 1$, there is a positive integer $k$ big enough so that $\frac1k<x$, and then $x\in\left(\frac1k,1\right]\subseteq\bigcup_{n=1}^\infty\left(\frac1n,1\right]$.

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$1)a.)$ $\bigcup_{n=1}^\infty [n,n+1] = [1,\infty)$,

$b.)$ $\bigcup_{n=1}^\infty (n,n+2) = (1,\infty)$.

$2) a.)$$\bigcup_{n=1}^\infty (n,n+1) = [1,\infty) \setminus \mathbb{N}$

$b.)$ $\bigcup_{n=1}^\infty \left(\dfrac{1}{n},1\right]=(0,1]$

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