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Prove:

$A \subset B \implies f^{-1}(A) \subset f^{-1}(B)$

I am busy setting up a proof for Real Analysis, and have come to a point where I need to use the above statement. Intuitively, I can draw a sketch of the above statement that quite clearly shows it to be true, but I would rather like to see an actual proof in order to know for certain that it is correct.

Can anyone please assist me in doing so?

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    $\begingroup$ Let $x\in f^{-1}(A)$; then $f(x)\in A$; therefore $f(x)\in B$. Hence $x\in f^{-1}(B)$. $\endgroup$ – egreg May 21 '15 at 18:02
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    $\begingroup$ This should be an answer. :) $\endgroup$ – fkraiem May 21 '15 at 18:02
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Whenever you have $$ x\in f^{-1}(A) $$ you can say $f(x)\in A$ and conversely. Thus $$ x\in f^{-1}(A)\implies f(x)\in A\implies f(x)\in B\implies x\in f^{-1}(B) $$

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  • $\begingroup$ Thank you! :). Since you posted first, I will accept your answer once the site allows me to :) $\endgroup$ – user860374 May 21 '15 at 18:06
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    $\begingroup$ I do not want to mess your nice answer, but I think you meant "you can say $f(x)\in A$" not $x\in A$, because $x \in f^{-1}(A)$. I can't correct it myself, since it is only 3 characters. $\endgroup$ – quapka May 21 '15 at 18:06
  • $\begingroup$ @quapka Typo, sorry! $\endgroup$ – egreg May 21 '15 at 19:34
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Take an arbitrary $x \in f^{-1}(A)$. By definition, $f(x) \in A$. As $A \subset B$, $f(x) \in B$. As $f(x) \in B$, it must also lie in ....

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