Introductory Real Analysis
Set Theory
Sets and Functions
To show $A=B$, show $A\subseteq B$ and $B\subseteq A$. Suppose $x\in B$, then we know by definition that $x\in (A\cup B)$ if $x\in B$ or $x\in A$. Which then implies that $x\in A$ from rule 1. Thus $B\subseteq A$.
Now suppose $x\in A$ which implies $x\in (A\cap B)$ (rule 2). The definition of this means that $x\in A$ and $x\in B$. $\therefore x\in A \implies x\in B$, i.e. $A\subseteq B$, so $A=B$.
This only holds for $B\subseteq A$. We proceed by counterexample.
Let $A={1,2}, B={3,4}$. Then $(A-B) = {1,2}$ and $(A-B)\cup B = {1,2,3,4} \neq {1,2}$.
Let $x\in (A-B)\cap C$. Then:
- $x\in A - B \implies x \in A $ and $x\not \in B$
- $x\in C$
So:
- $x \in A \cap C$
- Since $x\in C$ and $x\not\in B$, it follows that $x\not\in B \cap C$
Therefore $x\in (A\cap C)- (B\cap C)$
Let $x\in A \Delta B$. Then:
- $x\in A-B \implies x\in A$ and $x\not\in B$
OR:
- $x\in B-A \implies x\in B$ and $x\not\in A$
So
- $x\in A$ or $x \in B \implies x\in (A\cup B)$
- $x \not \in (A\cap B) \implies x\in (A\cap B)^c$
Therefore \[x\in(A\cup B) \cap (A\cap B)^c =(A\cup B)- (A\cap B)\]
Let $x\in \cup_\alpha A_\alpha - \cup_\alpha B_\alpha$. Then:
- $x\in \cup_\alpha A_\alpha \implies \exists \alpha{}_0: x \in A_\alpha{}_0$
- $x\not\in \cup_\alpha B_\alpha \implies \forall \alpha : x \not\in B_\alpha$
So
- $x\not\in B_\alpha{}_0$
Hence \[x\in A_\alpha{}_0 - B_\alpha{}_0 \subset \bigcup_\alpha\; (A_\alpha - B_\alpha)\]
Start with $I=\set{a,a+1}$, i.e. an arbitrary set of length 1. Then notice that you can subtract $a$ wlog ⊕ , and now we are tasked to find $\set{\langle x \rangle : x\in [0,1]}$. Furthermore, we know that $\langle x \rangle = x - \lfloor x \rfloor $ with $\langle 0 \rangle = 0 = \langle 1 \rangle$, whereby the image of the the closed interval only sweeps the half-open interval $[0,1)$.
$f$ cannot be one-to-one because of the periodicity; many real numbers have the same fractional parts.
The pre-image of $\frac{1}{4} \leq y \leq \frac{3}{4}$ is the interval $$ \bigcup_{n\in\mathbb{Z}} \left [\frac{1}{4} + n, \frac{3}{4} + n \right ]$$ because $x\in \mathbb{R} $.
Finally, we can express \[\mathbb{R} = \bigsqcup_{r\in [0,1)} \set{x\in \mathbb{R} : \langle x \rangle = r } \]
as the disjoint union of all the numbers which have the same fractional parts; i.e. the same images.
Equivalence of Sets. The Power of a Set
Ordered Sets and Ordinal Numbers
Systems of Sets
Metric Spaces
Basic Concepts
Let $(X,p)$ be a metric space. Prove that
- $\lvert p(x,z)-p(y,u)\rvert\le p(x,y)+p(z,u)$ for all $x,y,z,u\in X$;
- $\lvert p(x,z)-p(y,z)\rvert\le p(x,y)$ for all $x,y,z\in X$.
Deduce Minkowski’s integral inequality from Hölder’s inequality: \[ \Bigl(∫_a^b |f+g|\,p\Bigr)1/p ≤ \Bigl(∫_a^b |f|\,p\Bigr)1/p
\Bigl(∫_a^b |g|\,p\Bigr)1/p, \qquad p≥1. \]
Convergence. Open and Closed Sets
Complete Metric Spaces
Contraction Mappings
Topological Spaces
Basic Concepts
For any $M\subseteq T$ show that
- $[M]=M$ iff $M$ is closed;
- $[M]$ is the smallest closed set containing $M$;
- the closure operator satisfies Kuratowski’s axioms.
Let $\tau$ be the system consisting of $\varnothing$ and every subset of $X=[0,1]$ obtained by deleting a finite or countable set of points. Show that $(X,\tau)$ is
- not first‑countable,
- not second‑countable,
- a $T_1$ space but not Hausdorff.
Compactness
Real Functions on Metric and Topological Spaces