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Question Number 155586 by aaaspots last updated on 02/Oct/21

$$Howtoprooff:X\to Y$$$$fis1to1\Leftarrow \Rightarrow f\left(E\right)\mathrm{\setminus }f\left(F\right)=f\left(E\mathrm{\setminus }F\right)$$

Answered by mindispower last updated on 04/Oct/21

$$byDoubleinclusion$$$$letE,FbeesubsetoFX$$$$lety\in F\left(E\mathrm{\setminus }F\right)\Rightarrow \mathrm{\exists }!a\in E\mathrm{\setminus }Fsuchf\left(a\right)=y$$$$a\in E\mathrm{\setminus }F\Rightarrow f\left(a\right)\in f\left(E\right),f\left(a\right)\notin f\left(F\right)sincefisinjective$$$$\Rightarrow f\left(a\right)\in F\left(E\right)\mathrm{\setminus }F\left(F\right)\Rightarrow f\left(E\right)\mathrm{\setminus }f\left(F\right)\subseteq F\left(E\mathrm{\setminus }F\right)$$$$letz\in f\left(E\mathrm{\setminus }F\right)\Rightarrow \mathrm{\exists }!b\in E\mathrm{\setminus }F\mid f\left(b\right)=z$$$$\Rightarrow f\left(b\right)\in f\left(E\right),f\left(b\right)\notin f\left(F\right)becauseiff\left(b\right)\in f\left(F\right)$$$$supposef\left(b\right)\in f\left(F\right)\Rightarrow \mathrm{\exists }a\in Ff\left(a\right)=f\left(b\right)\Rightarrow a=bbyinjectivity$$$$a=b,b\in Fabsurdsinceb\in E\mathrm{\setminus }F$$$$\Rightarrow z\in f\left(E\right)\mathrm{\setminus }f\left(F\right)$$$$\Rightarrow f\left(E\mathrm{\setminus }F\right)\subseteq f\left(E\right)\mathrm{\setminus }f\left(F\right)$$$$\Rightarrow f\left(E\mathrm{\setminus }F\right)=f\left(E\mathrm{\setminus }F\right)$$$$$$

Commented by aaaspots last updated on 04/Oct/21

$$howabout1to1\Leftarrow f\left(E\mathrm{\setminus }F\right)=f\left(E\right)\mathrm{\setminus }f\left(F\right)$$

Commented by mindispower last updated on 04/Oct/21

$$nottrueifwetackF=\varnothing emptyset$$$$bydefinitionf\left(\varnothing \right)=\varnothing$$$$\Rightarrow \mathrm{\forall }f\in f(X\to Y)\mathrm{\forall }E\in P\left(X\right)$$$$fis1to1$$

Commented by aaaspots last updated on 07/Oct/21

$$afterthethirdlinesIstillunderstand$$$$$$

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