Lecture 8 - Review

This is just a review day, but I'm a bit under-prepared and haven't done these, so I'll be doing brief notes on hints towards selected problems:

1.5.4: Consider the construction we did in here. Now do the same construction for the interval $(a, b)$ rather than $(0, 1)$ . You get $f : (a, b) \to R$ $f (x) = \frac{x - \frac{a + b}{2}}{(x - a) (b - x)}$ . Then map values in $(a, b)$ to $(0, 1)$ and apply $f$ in the same way. Namely use $g (x) = \frac{x - a}{b - a}$ .
- You'll need to show surjectivity, so $\forall y \in R \exists x \in (0, 1) (f (x) = y)$ . Namely solve for $x$ to choose it:

\begin{aligned} y & = \frac{x - \frac{1}{2}}{x (1 - x)} \\ x (1 - x) y & = x - \frac{1}{2} \end{aligned}

Here if $y = 0$ then $x = \frac{1}{2}$ . If $y \neq 0$ then:

\begin{aligned} - y x^{2} + (y - 1) x + \frac{1}{2} & = 0 \\ x^{2} + \frac{1 - y}{y} x + \frac{1}{2 y} & = 0 \\ x & = \frac{\frac{y - 1}{y} \pm \sqrt{{(\frac{1 - y}{y})}^{2} - \frac{2}{y}}}{2} \end{aligned}

which is the $x$ you choose. For your proof start at the bottom $x = \dots$ and show that $f (y) = x$ as well as $x \in (0, 1)$ .

1.5.8: Let's do the problem:

Question

Let $B$ be a set of positive real numbers with the property that adding together any finite subset of elements from $B$ always gives a sum of $2$ or less. Show $B$ must be finite or countable.

Proof
Here $B \subseteq (0, \infty)$ where $\forall a, b \in B (a + b \leq 2)$ . Let's sketch it out:

The idea is that for any $n \in N$ then the interval $(0, \frac{1}{n})$ is still uncountable infinite.

Here $B \cap [2, \infty)$ has at most 1 element its' countable), ex $2$ . Also $B \cap [1, \infty)$ has finitely many elements. Repeat for $B \cap [\frac{1}{n}, \infty)$ all have finitely many elements. Note that:

⋃_{n = 1}^{\infty} [\frac{1}{n}, \infty) = {x \in R | x > 0}

Then $B = ⋃_{n = 1}^{\infty} (B \cap [\frac{1}{n}, \infty))$ so clearly we can count the finite union of subsets, so it is countable.

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Exam will be next Tuesday. Everything is fair game, but the proofs will be small and doable.