Question Number 155586 by aaaspots last updated on 02/Oct/21
$${How}\:{to}\:{proof}\:\:\:\:\:{f}:{X}\rightarrow{Y} \\ $$$${f}\:{is}\:\mathrm{1}\:{to}\:\mathrm{1}\:\Leftarrow\Rightarrow\:{f}\left({E}\right)\backslash{f}\left({F}\right)={f}\left({E}\backslash{F}\right) \\ $$
Answered by mindispower last updated on 04/Oct/21
$${by}\:{Double}\:{inclusion} \\ $$$${let}\:{E},{F}\:{bee}\:{subset}\:{oF}\:{X} \\ $$$${let}\:{y}\in{F}\left({E}\backslash{F}\right)\Rightarrow\exists!\:{a}\in{E}\backslash{F}\:{such}\:{f}\left({a}\right)={y} \\ $$$${a}\in{E}\backslash{F}\Rightarrow{f}\left({a}\right)\in{f}\left({E}\right),{f}\left({a}\right)\notin{f}\left({F}\right)\:{since}\:{f}\:{is}\:{injective} \\ $$$$\Rightarrow{f}\left({a}\right)\in{F}\left({E}\right)\backslash{F}\left({F}\right)\Rightarrow{f}\left({E}\right)\backslash{f}\left({F}\right)\subseteq{F}\left({E}\backslash{F}\right) \\ $$$${let}\:{z}\in{f}\left({E}\backslash{F}\right)\Rightarrow\exists!\:{b}\in{E}\backslash{F}\:\mid\:{f}\left({b}\right)={z} \\ $$$$\Rightarrow{f}\left({b}\right)\in{f}\left({E}\right),{f}\left({b}\right)\notin{f}\left({F}\right)\:\:\:\:{because}\:{if}\:{f}\left({b}\right)\in{f}\left({F}\right) \\ $$$${suppose}\:{f}\left({b}\right)\in{f}\left({F}\right)\Rightarrow\exists\:{a}\in{F}\:{f}\left({a}\right)={f}\left({b}\right)\Rightarrow{a}={b}\:{by}\:{injectivity} \\ $$$${a}={b},{b}\in{F}\:{absurd}\:{since}\:{b}\in{E}\backslash{F} \\ $$$$\Rightarrow{z}\in{f}\left({E}\right)\backslash{f}\left({F}\right) \\ $$$$\Rightarrow{f}\left({E}\backslash{F}\right)\subseteq{f}\left({E}\right)\backslash{f}\left({F}\right) \\ $$$$\Rightarrow{f}\left({E}\backslash{F}\right)={f}\left({E}\backslash{F}\right) \\ $$$$ \\ $$
Commented by aaaspots last updated on 04/Oct/21
$$\:\:\:{how}\:{about}\:\mathrm{1}\:{to}\:\mathrm{1}\Leftarrow\:\:{f}\left({E}\backslash{F}\right)={f}\left({E}\right)\backslash{f}\left({F}\right)\:\: \\ $$
Commented by mindispower last updated on 04/Oct/21
$${not}\:{true}\:{if}\:{we}\:{tack}\:{F}=\varnothing\:{empty}\:{set} \\ $$$${by}\:{definition}\:{f}\left(\varnothing\right)=\varnothing \\ $$$$\Rightarrow\forall\:{f}\in{f}\left({X}\rightarrow{Y}\right)\:\forall{E}\in{P}\left({X}\right) \\ $$$${f}\:{is}\:\mathrm{1}\:{to}\:\mathrm{1} \\ $$
Commented by aaaspots last updated on 07/Oct/21
$${after}\:{the}\:{third}\:{lines}\:{I}\:{still}\:{understand} \\ $$$$ \\ $$