12.11 Function

1. Consider the function \(\cos: \mathbb{R} \to [-1, 1]\) is bijective.

   1. Find an interval \([a, b]\) such that \(\cos: [a, b] \to [-1, 1]\) is bijective.
   2. Compute \(\cos^{-1}(y)\) for \(y \in \{-1, -\frac{1}{2}, 0, \frac{\sqrt{3}}{2}, 1\}\).
   3. Sketch the graph of \(\cos^{-1}\).

2. Let \(f: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) be given by \(f(x) = x + \frac{1}{x}\). Find:

   1. \(f^{-1}([0, 1])\)​
   2. \(f(f^{-1}(\mathbb{R}))\)​
   3. \(f^{-1}([3, 4])\)​

3. Let \(f: A \to B\) and let \(\{D_\alpha : \alpha \in \Delta\}\) be a family of subsets of \(A\).

   Prove that:

   1. \(f\left(\bigcap_{\alpha \in \Delta} D_\alpha\right) \subseteq \bigcap_{\alpha \in \Delta} f(D_\alpha)\)

      Let \(y \in f\left( \bigcap_{\alpha \in \Delta} D_\alpha \right)\), I will show that \(y \in \bigcap_{\alpha \in \Delta} f(D_\alpha)\).

      I know \(y = f(x)\) for \(x \in \bigcap_{\alpha \in \Delta} D_\alpha\), so \(x \in D_\alpha \ \forall \ \alpha \in \Delta\).

      So \(y = f(x)\) for \(x \in D_\alpha \ \forall \ \alpha \in \Delta\).

      \(\therefore y \in \bigcap_{\alpha \in \Delta} f(D_\alpha)\).

      What if \(\supseteq\)?

      Let's take \(y \in \bigcap_{\alpha \in \Delta} f(D_\alpha)\), then \(y\in f(D_{\alpha}),\forall\ \alpha\in\Delta\)

      Then there exists \(x_{\alpha}\in D_{\alpha}\ \forall\ \alpha\in\Delta\) such that \(y=f(x)\)​

      But i cannot ensure \(x_\alpha\) is the same one such that \(y=f(x_\alpha)\)​

      Thus this need injective 2. \(f\left(\bigcup_{\alpha \in \Delta} D_\alpha\right) = \bigcup_{\alpha \in \Delta} f(D_\alpha).\)

      \(\subseteq\)) Let \(y \in f\left(\bigcup_{\alpha \in \Delta} D_\alpha\right)\) \(\Rightarrow y = f(x)\) for \(x \in \bigcup_{\alpha \in \Delta} D_\alpha\) \(\Rightarrow x \in D_\alpha\) for some \(\alpha \in \Delta\) \(\Rightarrow y \in f(D_\alpha)\) for some \(\alpha \in \Delta\) \(\Rightarrow y \in \bigcup_{\alpha \in \Delta} f(D_\alpha)\)

      \(\supseteq\)) Let \(y \in \bigcup_{\alpha \in \Delta} f(D_\alpha)\) \(\Rightarrow y \in f(D_\alpha)\) for some \(\alpha \in \Delta\) \(\Rightarrow y = f(x)\) for \(x \in D_\alpha\) for some \(\alpha \in \Delta\) \(\Rightarrow x \in \bigcup_{\alpha \in \Delta} D_\alpha\)

4. Let \(f: A \to B\) be a function:

   Prove that \(X \subseteq f^{-1}(f(X))\), \(\forall X \subseteq A\).

   Prove that \(f(f^{-1}(Y)) \subseteq Y\), \(\forall Y \subseteq B\).

   Solution:

   1. Let \(x \in X \Rightarrow f(x) \in f(X)\). So \(x \in f^{-1}(f(X))\).

      How about \(\supseteq\) ?

      Take \(x\in f^{-1}(f(X))\), then \(x\in A:f(x)\in f(X)\), but we cannot imply \(x\in X\) since \(x\) may be in \(A\setminus X\) such that \(f(x)=y\) but \(x'\in X\) such that \(f(x')=y\)

      Thus \(f\) needs injective 2. \(y \in f(f^{-1}(Y)) \Rightarrow y = f(x)\) for \(x \in f^{-1}(Y)\). So \(y = f(x)\) for \(x\), an element such that \(f(x) \in Y\).
      So \(y \in Y\).

      How about \(\supseteq\)?

      Let take \(y\in Y\), then if \(\exists x\in A:f(x)=y\), then \(x\in f^{-1}(y)\). Then \(f(x)\in f(f^{-1}(y))\), then \(y\in f(f^{-1}(y))\subseteq f(f^{-1}(Y))\)​

      If \(f\) is not surjective in \(Y\), then doesn't hold

‍

‍