I am trying to describe $\bigcup_{i=1}^\infty A_i$ and $\bigcap_{i=1}^\infty A_i$ for the following two sets, but I am having a hard time conceptualizing it. $$(a)\ A_i = \{-i, -i+1,\ ...\ ,-1,0,1,\ ...\ ,i-1,i\}$$$$(b)\ A_i = \{-i, i\}$$ I think that for the set $(a)$, $\bigcup_{i=1}^\infty A_i$ represents any arbitrary set of integer, and $\bigcap_{i=1}^\infty A_i$ represents the set of all integers. However I'm not sure that this is correct, and I am having a hard time describing the union and intersection summations for $(b)$.

EDIT: formatting

  • $\begingroup$ What do you mean by represents any arbitrary set of integer? $\cup A_i$ is just one set. $\endgroup$ – mathcounterexamples.net Jun 30 '19 at 19:58
  • $\begingroup$ @mathcounterexamples.net I was saying that because I was treating $\cup$ like the $\lor$ operator. I'm still struggling to conceptualize this. $\endgroup$ – Robert Schwartz Jun 30 '19 at 20:06

Let’s denote in the two cases $A = \bigcup_{i=1}^\infty A_i$ and $B=\bigcap_{i=1}^\infty A_i$

(a) $\bigcup_{i=1}^\infty A_i = \mathbb Z$. Indeed any integer number $n$ belongs to $A_n$ and therefore to $A$, proving that $\mathbb Z \subseteq \bigcup_{i=1}^\infty A_i$. Conversely, every set $A_i$ is a set of integers, hence their union two. Therefore $\bigcup_{i=1}^\infty A_i \subseteq \mathbb Z$.

$\bigcap_{i=1}^\infty A_i = \{0\}$.

As $0$ belongs to all $A_i$ it belongs to their intersection. And none other integer belongs to this intersection. Which proves the equality.

(b) You have $\bigcup_{i=1}^\infty A_i = \mathbb Z \setminus \{0\}$ and $\bigcap_{i=1}^\infty A_i = \emptyset$. I left the proof to you.

  • $\begingroup$ This clears things up, thanks for the concise explanation. $\endgroup$ – Robert Schwartz Jun 30 '19 at 21:23

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