Show that $\{x\mid 0

Question

Show that $\{x\mid 0<x<2\}=A\subset B=\{x\mid 0<x<3\}$

I know it seems like a stupid question, it is even obvious, but, why is it true?

Here goes my take on it, but it is too direct and I don't think it is valid.

\begin{align} 0<a&\in A<2\\ 0<b&\in B<3 \end{align}

Because $0<a<2<3$ every value of $A$ will be in $B$.

Any thoughts of it? Take into account that I'm asking for the method to solve this, because $A$ could be something like $\{x\mid f(x)<x<g(x)\}$ where $f$ and $g$ are some functions.

Apply def $A \subseteq B$ iff : for all $x$, if $x \in A$, then $x \in B$. So: YES, your argument is fine : if $0 < x < 2$, then $0 < x < 3$. — Mauro ALLEGRANZA, Feb 24 '17 at 12:13
@MauroALLEGRANZA That is what i've done in $0<a<2<3$, right? — Garmekain, Feb 24 '17 at 12:15

score 1 · Accepted Answer · answered Feb 24 '17 at 12:15

1

If $x \in A$, then $0<x<2$. Since $2<3$, we get $0<x<3$, hence $x \in B$. That is all !

answered Feb 24 '17 at 12:15

Fred

77,394

I just needed to use better words for what I was thinking. – Garmekain Feb 24 '17 at 12:16

score 0 · Answer 2 · answered Feb 24 '17 at 12:27

In general, in order to show that a set $A$ is a subset of a set $B$, you can try the following argument: let $x \in A$, it means ..., let's show that $x \in B$.

Here: let $x \in A$, it means $0<x<2$. Since $2<3$, we have in fact $0<x<3$, that is $x \in B$.

Of course, it doesn't work every time, but it's the easiest way to think of inclusion.

Show that $\{x\mid 0

2 Answers2