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Show that $\{x\mid 0

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

Because $0

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)

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    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 $0 < x < 3$. – 2017-02-24
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    @MauroALLEGRANZA That is what i've done in $0 – 2017-02-24

2 Answers 2

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If $x \in A$, then $0

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    I just needed to use better words for what I was thinking. – 2017-02-24
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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

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