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I proved $\{0\}$ is contained in the intersection of $[0,x)$ where $0< x \leq 1$. But how do I show the reverse inclusion?

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closed as off-topic by M Turgeon, Andres Caicedo, Amzoti, Claude Leibovici, user127096 Apr 1 at 4:49

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Im sorry... but I dont know how to use matematical symbols in this app... T.T...... –  user138163 Apr 1 at 2:11
    
Oh ok I get it. You mean $\cap \{[0,x) : 0 < x \leq 1\}$. –  Guest Apr 1 at 2:13
    
Well, clearly the intersection is contained in $[0,1]$. If there was an $0 < y \leq 1$ such that $y$ was in the intersection, then you could find $0 < x < y$ for which $y \notin [0,x)$, a contradiction. Therefore the intersection is also contained in $\{0\}$. –  Guest Apr 1 at 2:14
    
@Nameless, I think you mean $[0,x) = \{y : 0 \leq y < x\}$. –  Antonio Vargas Apr 1 at 2:22
    
THANK YOU!!! Now I got it ! Thanks so much for your help :-) –  user138163 Apr 1 at 2:23

1 Answer 1

up vote 1 down vote accepted

Notice that $0\in[0,x),\forall x\in\mathbb{R}_{>0}$. Thus $\{0\}\subseteq \cap\{[0,x):0<x\leq 1\}$. Now we want to show no other number is in the intersection. Notice that $\forall c<0$, $c\not\in[0,x)\forall x$. For the other side, assume $\exists c>0$ such that $c\in\cap\{[0,x):0<x\leq 1\}$. Then $\forall x>0,c\in[0,x)$. But if $x=\frac{c}{2}$, then $c>x\Rightarrow c\not\in[0,x)$. Contradiction.

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Thank you so much for your help :) –  user138163 Apr 1 at 13:45
    
No problem. Don't forget to mark this as the answer by clicking the checkmark to the left of my answer. You can also up vote or down vote answers by pressing the up or down arrows. –  Joshua Biderman Apr 1 at 14:25
    
How kind you are. I checked and voted up your answer. Thank U !! –  user138163 Apr 1 at 14:33

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