Category: Set Theory

Question Number 1694 by 123456 last updated on 01/Sep/15 $$\mathrm{proof}\mathrm{that}\mathrm{for}\mathrm{two}\mathrm{set}\mathrm{A}\mathrm{and}\mathrm{B}\left(\mathrm{or}\mathrm{give}\mathrm{a}\mathrm{counter}\mathrm{example}\right)$$$$\mid \mathrm{A}\cup \mathrm{B}\mid ⩾\mid \mathrm{A}\cap \mathrm{B}\mid$$ Answered by Rasheed Soomro last updated on 01/Sep/15 $$\mathrm{In}\mathrm{the}\mathrm{following}\mathbf{A}\mathrm{and}\mathbf{B}\mathrm{are}\mathrm{assumed}\mathrm{as}\mathrm{finite}\mathrm{sets}.$$$$\:\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{B}}\:\subseteq\boldsymbol{\mathrm{A}}\subseteq\:\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\Rightarrow\mid\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{B}}\mid\leqslant\mid\boldsymbol{\mathrm{A}}\mid\leqslant\mid\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\mid………..\boldsymbol{\mathrm{I}}…

Question Number 1672 by 123456 last updated on 30/Aug/15 $$\mathrm{lets}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{be}\mathrm{two}\mathrm{finite}\mathrm{sets},\mathrm{proof}\left(\mathrm{or}\mathrm{give}\mathrm{a}\mathrm{counter}\mathrm{example}\right)\mathrm{that}$$$$\mid \mathrm{A}\cup \mathrm{B}\mid ⩽\mid \mathrm{A}\cap \mathrm{B}\mid \Rightarrow \mathrm{A}=\mathrm{B}$$ Answered by Rasheed Soomro last updated on 01/Sep/15 $$\mathbf{Case}-1\mathbf{W}\mathrm{hen}\mathbf{A}\cap \mathbf{B}=\varnothing ,\mid \mathbf{A}\cap \mathbf{B}\mid =0$$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{SubCase}}\:\:\left(\boldsymbol{\mathrm{a}}\right)\:\boldsymbol{\mathrm{At}}\:\boldsymbol{\mathrm{least}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{B}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{nonempty}}.…

Question Number 1132 by 123456 last updated on 24/Jun/15 $$\mathrm{let}\mathrm{S}=\{1,2,3,4,5\},\mathrm{if}\mathrm{A},\mathrm{B},\mathrm{C}\mathrm{is}\mathrm{such}\mathrm{that}$$$$\mathrm{A}\cap \mathrm{B}\cap \mathrm{C}=\varnothing$$$$\mathrm{A}\cap \mathrm{B}\ne \varnothing$$$$\mathrm{A}\cap \mathrm{C}\ne \varnothing$$$$\mathrm{how}\mathrm{many}\mathrm{ways}\mathrm{can}\mathrm{be}\mathrm{choose}\mathrm{A},\mathrm{B}\mathrm{and}$$$$\mathrm{C}$$ Commented by Faaiz…