The real numbers

The real numbers $\mathbb{R}$ satisfy three groups of axioms, and everything in analysis is deduced from them:

• Field axioms: $+$ and $\times$ are associative and commutative, $\times$ distributes over $+$, there are $0\ne1$, negatives $-x$, and inverses $x^{-1}$ for $x\ne0$.
• Order axioms: for any $x, y$ exactly one of $x < y$, $x = y$, $x > y$ holds; $<$ is transitive; $x < y\Rightarrow x + z < y + z$; and $x, y > 0\Rightarrow xy > 0$.
• Completeness: every non-empty subset of $\mathbb{R}$ that is bounded above has a least upper bound (its supremum; see Prop. 1.3). $\mathbb{Q}$ satisfies the first two groups but not completeness: $\{q\in\mathbb{Q}: q^2 < 2\}$ has no supremum in $\mathbb{Q}$.

The absolute value is $|x| = x$ if $x\ge0$ and $|x| = -x$ if $x < 0$; $|x - a|$ is the distance from $x$ to $a$, and $|x - a| < \varepsilon\iff a - \varepsilon < x < a + \varepsilon$. The interval $(a - \varepsilon, a + \varepsilon)$ is the $\varepsilon$-neighbourhood of $a$. The working tools: $|xy| = |x||y|$; the triangle inequality $|x + y|\le|x| + |y|$ (check the cases $x + y\ge0$ and $x + y < 0$); its reverse form $\big||x| - |y|\big|\le|x - y|$. Key fact: if $|x - y| < \varepsilon$ for every $\varepsilon > 0$ then $x = y$.

For $S\ne\emptyset$ bounded above, $u = \sup S$ iff (i) $s\le u$ for all $s\in S$, and (ii) for every $\varepsilon > 0$ there is $s\in S$ with $s > u - \varepsilon$. Hence there is a sequence $s_n\in S$ with $s_n\to\sup S$. Similarly for $\inf$, and $\inf S = -\sup(-S)$. Order properties: if $S\subseteq T$ then $\sup S\le\sup T$; if $s\le t$ for all $s\in S, t\in T$ then $\sup S\le\inf T$.

For every $x\in\mathbb{R}$ there is $n\in\mathbb{N}$ with $n > x$. Equivalently, for every $\varepsilon > 0$ there is $N$ with $\frac1N < \varepsilon$, and then $\frac1n < \varepsilon$ for all $n\ge N$. This is the fact behind $\frac1n\to0$, and behind every "choose $N > 1/\varepsilon$" in an $\varepsilon$–$N$ proof.

Between any two real numbers $a < b$ there is a rational and an irrational number. Consequently every real number is a limit of a sequence of rationals and of a sequence of irrationals, and every interval of positive length contains infinitely many of each.

If $I_n = [a_n, b_n]$ are closed bounded intervals with $I_1\supseteq I_2\supseteq\cdots$, then $\bigcap_nI_n\ne\emptyset$. If also $b_n - a_n\to0$, the intersection is a single point. Both "closed" and "bounded" matter: $\bigcap(0, \frac1n) = \emptyset$ and $\bigcap[n, \infty) = \emptyset$.

To show $u = \sup S$: (1) prove $s\le u$ for every $s\in S$; (2) given any $\varepsilon > 0$, produce an explicit $s\in S$ with $s > u - \varepsilon$, usually by choosing $n$ large via the Archimedean property. Say whether $u\in S$ (then it's a maximum). Common slips: proving only (1); forgetting $S$ might attain its sup at a small $n$; using a limit you haven't justified.