# The intersection of $[0,x)$ where $0<x\leq 1$ is $\{0\}$ [closed]

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, Andrés E. Caicedo, Amzoti, Claude Leibovici, user127096Apr 1 '14 at 4:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – Andrés E. Caicedo, user127096
If this question can be reworded to fit the rules in the help center, please edit the question.

Im sorry... but I dont know how to use matematical symbols in this app... T.T...... – user138163 Apr 1 '14 at 2:11
Oh ok I get it. You mean $\cap \{[0,x) : 0 < x \leq 1\}$. – Guest Apr 1 '14 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 '14 at 2:14
@Nameless, I think you mean $[0,x) = \{y : 0 \leq y < x\}$. – Antonio Vargas Apr 1 '14 at 2:22
THANK YOU!!! Now I got it ! Thanks so much for your help :-) – user138163 Apr 1 '14 at 2:23

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.