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  • MAT102 Final Exam Prep

MAT102 Worksheet 10 Review

#r# Q1

• Suppose $f:A\to B$ is a function

A

• Show that if $U_1,U_2\subseteq B$ then $f^{-1}(U_1\cap U_2)=f^{-1}(U_1)\cap f^{-1}(U_2)$
• Double Subset Inclusion.
• $\subseteq$
  • let $x\in f^{-1}(U_1\cap U_2)$
  • So $x\in f^{-1}(U_1)$ and $x\in f^{-1}(U_2)$
  • So then $x\in f^{-1}(U_1)\cap f^{-1}(U_2)$
• $\supseteq$
  • Let $x\in f^{-1}(U_1)\cap f^{-1}(U_2)$
  • So $x\in f^{-1}(U_1)$ and $f^{-1}(U_2)$
  • Then $x\in f^{-1}(U_1\cap U_2)$

B

• Show that if $U_1,U_2,\dots ,U_n\subseteq B$ then:
  • $f^{-1}\left(\bigcap _{i=1}^n{U}_i\right)=\bigcap _{i=1}^n{f}^{-1}(U_i)$
• Base Case: $n=1$
  • This was proven in part A
• Induction Hypothesis:
  • For some arbitrary $n\in \mathbb{N}$, $f^{-1}\left(\bigcap _{i=1}^n{U}_i\right)=\bigcap _{i=1}^n{f}^{-1}(U_i)$ is true.
• Induction Step:
  • We want to show that $f^{-1}\left(\bigcap _{i=1}^n{U}_i\right)=\bigcap _{i=1}^n{f}^{-1}(U_i)⟹f^{-1}\left(\bigcap _{i=1}^{n+1}{U}_i\right)=\bigcap _{i=1}^{n+1}{f}^{-1}(U_i)$
  • Notice that $f^{-1}\left(\bigcap _{i=1}^{n+1}{U}_i\right)=f^{-1}(\bigcap _{i=1}^n{U}_i\cap U_{n+1})$
  • Notice that $\bigcap _{i=1}^{n+1}{f}^{-1}(U_i)=\bigcap _{i=1}^n{f}^{-1}(U_i)\cap f^{-1}(U_{n+1})$
  • $f^{-1}(\bigcap _{i=1}^n{U}_i\cap U_{n+1})=\bigcap _{i=1}^n{f}^{-1}(U_i)\cap f^{-1}(U_{n+1})$
  • $\subseteq$
    • Let $x\in f^{-1}(\bigcap _{i=1}^n{U}_i\cap U_{n+1})$
    • So $x\in f^{-1}\left(\bigcap _{i=1}^n{U}_i\right)$ and $x\in f^{-1}(U_{n+1})$
    • So then by part A we know that $x\in \bigcap _{i=1}^n{f}^{-1}(U_i)$ and $x\in f^{-1}(U_{n+1})$
    • So then $x\in \bigcap _{i=1}^n{f}^{-1}(U_i)\cap f^{-1}(U_{n+1})$
    • So $x\in \bigcap _{i=1}^{n+1}{f}^{-1}(U_i)$
  • $\supseteq$
    • Let $x\in \bigcap _{i=1}^n{f}^{-1}(U_i)\cap f^{-1}(U_{n+1})$
    • So $x\in \bigcap _{i=1}^n{f}^{-1}(U_i)$ and $f^{-1}(U_{n+1})$
    • By part A
    • $x\in f^{-1}\left(\bigcap _{i=1}^n{U}_i\right)$ and $x\in f^{-1}(U_{n+1})$
    • So then again by part a $x\in f^{-1}(\bigcap _{i=1}^n{U}_i\cap U_{n+1})$
    • Finally $x\in f^{-1}\left(\bigcap _{i=1}^{n+1}{U}_i\right)$
  • $◻$