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 $(1, 2, \ldots, \infty)$

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


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


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


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


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 $x\in\left(\frac1k,1\right]\subseteq\bigcup_{n=1}^\infty\left(\frac1n,1\right]$.


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

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

$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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