Math Challenges

Submissions for Problem #68

Problem #68

\[ \displaylines{\text{Let }X\text{ be a set, and let }A\subseteq X.\text{ Define a function }f:P\left(x\right)\to P\left(X\right)\text{ by }f\left(B\right)=A\cap B,\text{ for }B\in P\left(X\right).\\ \text{Under what conditions is }f\text{ one-to-one and onto?}} \]

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zapwai

Solution: \( \displaylines{f\left(\empty\right)=\empty\text{ implies that }f\text{ is not one-to-one if }B=A^{c}\text{ (as }f\left(A^{c}\right)=A\cap A^{c}=\empty).\\ \text{Therefore }A=X\text{ if }f\text{ is to be one-to-one, and this also makes }f\text{ an onto map.}\\ } \)