MAT102 Lecture 17

• Functions in Math
• These are diff functions:
  • $f:\mathbb{R}\to \mathbb{R},f(x)=x^2$
  • $g:\mathbb{R}\to [0,\mathrm{\infty }),g(x)=x^2$
  • $g$ has a right inverse but $f$ doesn't have an inverse.
  • Even though the graphs are the same, the 'function' is different.
• $f:\{0,1\}\to \mathbb{R},f(x)=x^2$
• $g:\{0,1\}\to \mathbb{R},g(x)=x$
• These are the same function, why? Because the input is the same, the outputs are also the same.
  • $f(0)=0^2=0$
  • $f(1)=1^2=1$
  • $g(0)=0$
  • $g(1)=1$
  • So you can change the definition of how to compute the outputs - in certain cases you can have the exact same function then.
• The average function from $\mathbb{R}\to \mathbb{R}$ is not continuous: probability is 0.
• The average function from $\mathbb{R}\to \mathbb{R}$ if it's continuous, is not differentiable everywhere: probability is 0.
• The Weierstrass Function is basically a triangle wave distorted with some white noise. It's also one of the only always continuous and differentiable function.
• A lambda is like a mini function:
  • In python, you can do f=lambda x: x*x
  • In math it's in functions, when you see $x\mapsto x^2$ it means $g(x)=x^2$
• Invertible Functions in Math
• $f:\mathbb{R}\to \mathbb{R},f(x)=x^2$
• The preimage of this:
  • $f^{-1}(\mathbb{R})=[0,\mathrm{\infty })$
  • Preimage of the Codomain
  • This is the domain, everything in the Domain maps to the Codomain
  • $f(f^{-1}(\mathbb{R}))=f(\mathbb{R})=[0,\mathrm{\infty })$

• 1 Suppose that $f:A\to B$ is a function
  • a
    • 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
    • $⟹$
      • If $x\in f^{-1}(U_1\cap U_2)$
      • This means $f(x)\in U_1$ and $f(x)\in U_2$
      • $x\in f^{-1}(U_1)$ and $x\in f^{-1}(U_2)$
      • So this means $x\in f^{-1}(U_1)\cap f^{-1}(U_2)$
    • $⟸$
      • If $x\in f^{-1}(U_1)\cap f^{-1}(U_2)$
      • This means $x\in f^{-1}(U_1)$ and $x\in f^{-1}(U_2)$
      • So $f(x)\in U_1$ and $f(x)\in U_2$
      • So $f(x)\in U_1\cap U_2$
      • So $x\in f^{-1}(U_1\cap U_2)$
  • 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: $N=2$
      • This is true theoretically because the image is the range. The range will be in the Codomain.
      • $f^{-1}(U_1\cap U_1)=f^{-1}(U_1)\cap f^{-1}(U_1)$
      • This was proven in 1)a)
    • Induction Hypothesis:
      • For $k\ge 2$
      • $f^{-1}\left(\bigcap _{i=1}^k{U}_i\right)=\bigcap _{i=1}^k{f}^{-1}(U_i)$
    • Induction Step:
      • Show that this is true for $n=k+1$
      • $f^{-1}\left(\bigcap _{i=1}^{k+1}{U}_i\right)=\bigcap _{i=1}^{k+1}{f}^{-1}(U_i)$
      • $f^{-1}(\bigcap _{i=1}^k{U}_i\cap U_{k+1})=\bigcap _{i=1}^{k+1}{f}^{-1}(U_i)$
      • $f^{-1}(\bigcap _{i=1}^k{U}_i\cap U_{k+1})=\bigcap _{i=1}^k{f}^{-1}(U_i)\cap f^{-1}(U_{k+1})$
      • We see the induction hypothesis on the right side
      • $f^{-1}(U_1\cap U_2\cap \cdots \cap U_{k+1})=\bigcap _{i=1}^k{f}^{-1}(U_i)\cap f^{-1}(U_{k+1})$
  • Solutions
    • a
      • $\subseteq$
        • Let $x\in f^{-1}(U_1\cap U_2)$
        • So $f(x)\in U_1\cap U_2$
        • So $f(x)\in U_1$ and $f(x)\in U_2$
        • So $x\in f^{-1}(U_1)$ and $x\in f^{-1}(U_2)$
        • So $x\in f^{-1}(U_1)\cap f^{-1}(U_2)$
        • This is simple
      • $\supseteq$
        • …
    • b
      • Base Case: $n=2$
        • $f^{-1}(U_1\cap U_2)=f^{-1}(U_1)\cap f^{-1}(U_2)$
        • Part 1)a)
      • Assume for some $n$, $f^{-1}\left(\bigcap _{i=1}^n{U}_i\right)=\bigcap _{i=1}^n{f}^{-1}(U_i)$
      • $f^{-1}\left(\bigcap _{i=1}^{n+1}{U}_i\right)=f^{-1}(\bigcap _{i=1}^n{U}_i\cap U_{n+1})=f^{-1}\left(\bigcap _{i=1}^n{U}_i\right)\cap f^{-1}(U_{n+1})$ We have part 1)a)
      • $\bigcap _{i=1}^n{f}^{-1}(U_i)\cap f^{-1}(U_{n+1})$ Induction hypothesis
      • $\bigcap _{i=1}^{n+1}{f}^{-1}(U_i)$
    • c
      • 1
        • Let $I_k=[-\frac{1}{k},\frac{1}{k}]\subseteq \mathbb{R}$ for $k$ a positive integer
        • Determine $\bigcap _{k=1}^{\mathrm{\infty }}{I}_k$
        • Claim: The $\bigcap _{k=1}^{\mathrm{\infty }}[-\frac{1}{k},\frac{1}{k}]=\{0\}$
        • Double Subset
        • $\supseteq$
          • Note that $0\in [-\frac{1}{k},\frac{1}{k}]$ for all $k$, so
          • $0\in \bigcap _{k=1}^{\mathrm{\infty }}{I}_k$
        • $\subseteq$
          • Let $x\in \bigcap _{k=1}^{\mathrm{\infty }}{I}_k$
            • Show that $x=0$
          • Contradiction, suppose it's not
            • $x>0$
            • Thus there is some $k\in \mathbb{N}$ such that $\frac{1}{k}<x$
            • Therefore $x\notin [-\frac{1}{k},\frac{1}{k}]$ so
            • $x\notin \bigcap _{k=1}^{\mathrm{\infty }}[-\frac{1}{k},\frac{1}{k}]$, a contradiction.
      • 2
        • basically part a