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Question

# For any two sets A and B, show that the following statements are equivalent: (i) $A\subset B$ (ii) $A-B=\varphi$ (iii) $A\cup B=B$ (iv) $A\cap B=A.$

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Solution

## We have that the following statements are equivalent: (i) $A\subset B$ (ii) $A-B=\varphi$ (iii) $A\cup B=B$ (iv) $A\cap B=A$ Proof: $\mathrm{Let}A\subset B\phantom{\rule{0ex}{0ex}}\mathrm{Let}x\mathrm{be}\mathrm{an}\mathrm{arbitary}\mathrm{element}\mathrm{of}\left(A-B\right).\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}x\in \left(A-B\right)\phantom{\rule{0ex}{0ex}}⇒x\in A&x\notin B\left(\mathrm{Which}\mathrm{is}\mathrm{contradictory}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Also},\phantom{\rule{0ex}{0ex}}\because A\subset B\phantom{\rule{0ex}{0ex}}⇒A-B\subseteq \mathrm{\varphi }...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{null}\mathrm{sets}\mathrm{are}\mathrm{the}\mathrm{subsets}\mathrm{of}\mathrm{every}\mathrm{set}.\phantom{\rule{0ex}{0ex}}\therefore \mathrm{\varphi }\subseteq A\mathit{-}B\mathit{}...\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right)&\left(2\right),\mathrm{we}\mathrm{get},\phantom{\rule{0ex}{0ex}}\left(A\mathit{-}B\right)=\mathrm{\varphi }\phantom{\rule{0ex}{0ex}}\therefore \left(\mathrm{i}\right)=\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{we}\mathrm{have},\phantom{\rule{0ex}{0ex}}\left(A\mathit{-}B\right)=\mathrm{\varphi }\phantom{\rule{0ex}{0ex}}\mathrm{That}\mathrm{means}\mathrm{that}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{element}\mathrm{in}A\mathrm{that}\mathrm{does}\mathrm{not}\mathrm{belong}\mathrm{to}B.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathit{}A\cup B\mathit{=}B\phantom{\rule{0ex}{0ex}}\therefore \left(\mathrm{ii}\right)=\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}A\cup B=B\phantom{\rule{0ex}{0ex}}⇒A\subset B\phantom{\rule{0ex}{0ex}}⇒A\cap B\mathit{=}A\phantom{\rule{0ex}{0ex}}\therefore \left(\mathrm{iii}\right)=\left(\mathrm{iv}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}A\cap B\mathit{=}A\phantom{\rule{0ex}{0ex}}\mathrm{It}\mathrm{should}\mathrm{be}\mathrm{possible}\mathrm{if}\mathit{}A\subset B.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}A\subset B\phantom{\rule{0ex}{0ex}}\therefore \left(\mathrm{iv}\right)=\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)=\left(\mathrm{ii}\right)=\left(\mathrm{iii}\right)=\left(\mathrm{iv}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}\mathrm{all}\mathrm{statements}\mathrm{are}\mathrm{equivalent}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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