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If $S$ is the set of real numbers, and if $T$ is the set of rational numbers, let, for $\alpha \in T, \ A_{\alpha}= \{x\in S\ | \ x\ge \alpha\}$.

Can anyone help explain why $\cup_{\alpha\in T}A_{\alpha}=S$ and $\cap_{\alpha\in T}A_{\alpha} =\emptyset$? Thanks!

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Try and interpret the following in terms of unions and intersections, and the sets $A_{\alpha}$ you're given:

  • Given any real number, it lies above some rational number. (This proves the first equation.)

  • No real number lies above all rational numbers. (This proves the second equation.)

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  • $\begingroup$ So that is what $A_{\alpha}= \{x\in S\ | \ x\ge \alpha\}$ is saying? That all real numbers are greater than rational numbers. I think I got confused of $A_{\alpha}$ since $\alpha \in T$ so I didn't know what it meant. You think you can tell me what $A_{\alpha}= \{x\in S\ | \ x\ge \alpha\}$ is saying? $\endgroup$
    – Tom
    Aug 22 '13 at 1:54
  • $\begingroup$ Given a rational number $\alpha$, the set $A_{\alpha}$ is the set of real numbers $x$ for which $x \ge \alpha$. That's what the notation means. $\endgroup$ Aug 22 '13 at 2:00
  • $\begingroup$ ...so the union is the collection of real numbers $x$ for which there exists a rational number $\alpha$ with $x \ge \alpha$ (i.e. the set of reals greater than some rational number); and the intersection is the collection of real numbers $x$ for which $x \ge \alpha$ for all rational numbers $\alpha$ (i.e. the set of reals greater than all rational numbers). HTH. $\endgroup$ Aug 22 '13 at 2:01
  • $\begingroup$ I got it now with your help. Thanks!! $\endgroup$
    – Tom
    Aug 22 '13 at 2:26
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This follows from the fact that the rational numbers are unbounded both from above and from below, in the set of real numbers.

Recall the definitions:

$$\bigcup_{i\in I}X_i=\{x\mid\exists i\in I\text{ such that}x\in X_i\}\\ \bigcap_{i\in I}X_i=\{x\mid\forall i\in I\text{ such that}x\in X_i\}$$

Apply these two definitions to the ones in your problem, and use the fact from above to see the equalities hold.

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  • $\begingroup$ I think what I am having difficulty understanding is what $A_{\alpha}= \{x\in S\ | \ x\ge \alpha\}$ is saying exactly. What I got from that is, for all $A_{\alpha}$ where $\alpha$ is a rational number there is an $x\in S$ such that $x\ge$ rational numbers. $\endgroup$
    – Tom
    Aug 22 '13 at 1:57
  • $\begingroup$ Tom, $A_\alpha$ is simply the set of those real numbers which are either equal to $\alpha$ or larger than $\alpha$. For example for $\alpha=0$ this means all the positive real numbers, and $0$ itself. $\endgroup$
    – Asaf Karagila
    Aug 22 '13 at 2:04
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Sorry but

$\cup_{\alpha\in T}A_{\alpha}=S$ is FALSE!

Let S = {2,3} and T = {5} then in this case

$\cup_{\alpha\in T}A_{\alpha}=$ {x∈{2,3} | x≥5} = ∅ because neither 2 nor 3 are greater than 5 and S = {2,3} is not equal to ∅

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  • $\begingroup$ $S$ is the set of real number, not an arbitrary set so this does not answer the question. $\endgroup$
    – Winther
    Sep 30 '15 at 0:01

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