MT2025 Analysis I lecture notes

Proof. As $𝑎+𝑥=𝑎$, then

By A2,

Hence $𝑥=0$.⁠$\square$

Proof.

so $𝑎\times 0=0$ by additive inverse.

⁠$\square$

Proof. By contraposition.

Suppose $𝑎\times 𝑏=0$ and $𝑎\ne 0$.

Then

By M2, M1, M4, LHS $=1\times 𝑏=0$, so by M3, $𝑏=0$.⁠$\square$

Proof.

$\le$ is reflexive: $𝑎-𝑎=0\in \mathbb{P}\cup \left\{0\right\}$, so $𝑎\le 𝑎$ for all $𝑎\in ℝ$.

$\le$ is antisymmetric: suppose $𝑎\le 𝑏$ and $𝑏\le 𝑎$. Then $𝑏-𝑎\in \mathbb{P}\cup \left\{0\right\}$ and $-\left(𝑏-𝑎\right)=𝑎-𝑏\in \mathbb{P}\cup \left\{0\right\}$. These contradict P3 unless $𝑏-𝑎=0⟹𝑎=𝑏$.

$\le$ is transitive: suppose $𝑎\le 𝑏$ and $𝑏\le 𝑐$. Then $𝑏-𝑎,𝑐-𝑏\in \mathbb{P}\cup \left\{0\right\}$. If a pair of $𝑎,𝑏,𝑐$ are equal, clearly $𝑎\le 𝑐$. Otherwise, $𝑏-𝑎,𝑐-𝑏\in \mathbb{P}$ so by P1, $𝑐-𝑎=\left(𝑐-𝑏\right)+\left(𝑏-𝑎\right)\in \mathbb{P}$ so $𝑎\le 𝑐$.

Every pair $𝑎,𝑏\in ℝ$ is comparable:

• If $𝑎=𝑏$, done
• If $𝑎<𝑏$, done
• By trichotomy, the only other option is $𝑎-𝑏=-\left(𝑏-𝑎\right)\in \mathbb{P}$, so $𝑏<𝑎$ (so $𝑏\le 𝑎$).⁠$\square$
  

(i). $\left(𝑏+𝑐\right)-\left(𝑎+𝑐\right)=𝑏-𝑎\in \mathbb{P}\cup \left\{0\right\}$ by assumption.⁠$\square$

(ii).

$𝑏𝑐-𝑐𝑎=𝑐\left(𝑏-𝑎\right)$.

If $𝑐=0$ or $𝑏=𝑎$, then $𝑐\left(𝑏-𝑎\right)\in \mathbb{P}\cup \left\{0\right\}$.

If $𝑐>0$ and $𝑏>𝑎$, then $𝑐\left(𝑏-𝑎\right)\in \mathbb{P}$ by P2.⁠$\square$

Proof. Suppose for the sake of contradiction that there is such a $\mathbb{P}$.

By P3, one of $𝑖\in \mathbb{P},𝑖=0,-𝑖\in \mathbb{P}$.

If $𝑖=0$, then $1={𝑖}^4=0^4=0$, contradicting axiom Z.

If $𝑖\in \mathbb{P}$, then by P2, $-𝑖={𝑖}^3\in \mathbb{P}$. This contradicts P3.

If $-𝑖\in \mathbb{P}$, we get the same contradiction.

Therefore $ℂ$ is not an ordered field.⁠$\square$

Proof. For $𝑥,𝑦\in \left[0,\infty \right)$ s.t. $𝑥\le 𝑦$, ${𝑦}^2-{𝑥}^2=\left(𝑦-𝑥\right)\left(𝑦+𝑥\right)\in \left[0,\infty \right]$ by P2. This satisfies the definition of ${𝑥}^2\le {𝑦}^2$, so $𝑥\mapsto {𝑥}^2$ is increasing on this interval.⁠$\square$

Proof.

By Lemma 1.2.7, $\left|𝑎+𝑏\right|\le \left|𝑎\right|+\left|𝑏\right|$.

⁠$\square$

Proof. By induction.

When $𝑛=0$, LHS $=1=$ RHS.

Suppose true for $𝑛$. Then

Hence true by induction.

⁠$\square$

Proof.

1. Suppose $𝑎,𝑏$ are maxima of $𝑆$. Then $𝑎\ge 𝑏$ as $𝑎$ is a max and $𝑏\in 𝑆$. Also $𝑏\ge 𝑎$ by the same argument. Hence $𝑎=𝑏$.
2. A least upper bound is an upper bound which is less than or equal to any upper bound. For $𝑎,𝑏$ two least upper bounds, $𝑎\le 𝑏$ and $𝑏\le 𝑎$ so $𝑎=𝑏$.
3. $max𝑆\le sup𝑆$ because $max𝑆\in 𝑆$. But $max𝑆$ is an upper bound itself, so $sup𝑆\le max𝑆$. Therefore $max𝑆=sup𝑆$.

⁠$\square$

Proof. If there were no such $𝑠\in 𝑆$, then $sup𝑆-𝜀$ would be an upper bound of $𝑆$ less than the least upper bound, which is a contradiction.⁠$\square$

Proof. Let $𝑆=\left\{𝑥\in ℝ|{𝑥}^2\le 2\right\}$.

Note that $1\in 𝑆$ and that 2 bounds $𝑆$ from above. We set $𝛼=sup𝑆$, which exists by the completeness axiom.

By trichotomy, ${𝛼}^2=2$ or ${𝛼}^2>2$ or ${𝛼}^2<2$.

Suppose for the sake of contradiction that ${𝛼}^2>2$. Pick $0<ℎ<\frac{{𝛼}^2-2}{2𝛼}$, then ${\left(𝛼-ℎ\right)}^2>2$. Then no element of $𝑆$ lies in $\left(𝛼-ℎ,𝛼\right)$. This contradicts the approximation property.

Suppose for the sake of contradiction that ${𝛼}^2<2$. Pick $ℎ<1$ (so that ${ℎ}^2<ℎ$), then

giving ${\left(𝛼+ℎ\right)}^2<2$. Therefore $𝛼+ℎ\in 𝑆$, but $𝛼+ℎ>𝛼=sup𝑆$, contradiction.

By trichotomy, ${𝛼}^2=2$.

To show uniqueness, suppose that $𝛽>0$ and ${𝛽}^2=2$. Then ${𝛼}^2={𝛽}^2=2$ and

Since $𝛼+𝛽>0$, it must be that $𝛼-𝛽=0$ so $𝛼=𝛽$.⁠$\square$

Proof. Suppose for the sake of contradiction that such an upper bound exists. As $\mathbb{N}\ne \varnothing$, we can choose $𝛼=sup\mathbb{N}$. By the approximation property with $𝜀=1$, there exists $𝑛\in \mathbb{N}$ such that $𝛼-1<𝑛\le 𝛼$. But $𝑛+1\in \mathbb{N}$ and $𝑛+1>𝛼=sup\mathbb{N}$. Contradiction.⁠$\square$

(by Cantor). We put ${\mathbb{Q}}^{+}$ in a grid indexed by numerator and denominator. We can count by the elements of ${\mathbb{Q}}^{\ast }$ by zigzagging diagonally, omitting repetitions.

There’s then a natural bijection to $\mathbb{Q}$.⁠$\square$

Proof. We use the same diagonal counting as for $\mathbb{Q}$.⁠$\square$

Proof. We show that $\left[0,1\right]$ is uncountable.

Suppose for the sake of contradiction that we can list the elements of $\left[0,1\right]$ by their decimal expansions, such that ${𝑟}_0=0.{𝑎}_{01}{𝑎}_{02}{𝑎}_{03}\dots$, ${𝑟}_1=0.{𝑎}_{11}{𝑎}_{12}{𝑎}_{13}\dots$ etc.

Then we can construct a new number $𝑥\in \left[0,1\right]$ different from each ${𝑟}_{𝑖}$, by making sure the $𝑖$th digit place of $𝑥$ is different from the $𝑖$th digit of ${𝑟}_{𝑖-1}$, for instance by choosing ${𝑥}_{𝑖}=7$ if ${𝑎}_{\left(𝑖-1\right)𝑖}\ne 7$, and 6 otherwise.

By construction, $𝑥\ne {𝑟}_{𝑖}$ for all $𝑖\in \mathbb{N}$, so the enumeration was not exhaustive. Contradiction.⁠$\square$

Proof. Suppose for the sake of contradiction that ${𝑥}_{𝑛}\to {𝐿}_1$ and ${𝑥}_{𝑛}\to {𝐿}_2$, for ${𝐿}_1\ne {𝐿}_2$.

Let $𝜀≔\frac{1}{2}\left|{𝐿}_1-{𝐿}_2\right|>0$.

Since ${𝑥}_{𝑛}\to {𝐿}_1$, there is a tail of $\left({𝑥}_{𝑛}\right)$ in $\left({𝐿}_1-𝜀,{𝐿}_1+𝜀\right)$. Say that this is the ${𝑁}_1$th tail.

Similarly, there is an ${𝑁}_2$ such that the ${𝑁}_2$th tail of $\left({𝑥}_{𝑛}\right)$ is in $\left({𝐿}_2-𝜀,{𝐿}_2+𝜀\right)$.

W.L.O.G. say that ${𝐿}_1<{𝐿}_2$. Then ${𝐿}_1+𝜀=\frac{{𝐿}_1+{𝐿}_2}{2}$ and ${𝐿}_2-𝜀=\frac{{𝐿}_1+{𝐿}_2}{2}$.

For $𝑛\ge max\left({𝑁}_1,{𝑁}_2\right)$, we have ${𝑥}_{𝑛}<\frac{{𝐿}_1+{𝐿}_2}{2}<{𝑥}_{𝑛}$. Contradiction.⁠$\square$

Proof. Set $𝜀=1$. As ${𝑥}_{𝑛}\to 𝐿$, there exists $𝑁\in \mathbb{N}$ such that, for all $𝑛\ge 𝑁$, $\left|{𝑥}_{𝑛}-𝐿\right|<1$.

For $𝑛\ge 𝑁$, $\left|{𝑥}_{𝑛}\right|<\left|𝐿\right|+1$ by the triangle inequality.

So for all $𝑛$, we have $\left|{𝑥}_{𝑛}\right|\le max\left\{\left|𝐿\right|+1,\left|{𝑥}_1\right|,\left|{𝑥}_2\right|,\dots ,\left|{𝑥}_{𝑁-1}\right|\right\}$.

Hence $\left({𝑥}_{𝑛}\right)$ is bounded.⁠$\square$

Proof. Suppose $𝑛>𝑘$, and write $𝑎=1+𝑏$.

By the binomial theorem,

Therefore

Given any $𝜀>0$ we can choose suitably large $𝑁$ such that

for all $𝑛\ge 𝑁$.⁠$\square$

Proof. Suppose that ${𝑥}_{𝑛}\to {𝐿}_1$ and ${𝑦}_{𝑛}\to {𝐿}_2$, and let $𝐿={𝐿}_1+𝑖{𝐿}_2$.

Let $𝜀>0$. As ${𝑥}_{𝑛}\to {𝐿}_1$, $\exists {𝑁}_1$ s.t. $\forall 𝑛\ge {𝑁}_1$, $\left|{𝑥}_{𝑛}-{𝐿}_1\right|<\frac{𝜀}{2}$. Similarly, $\exists {𝑁}_2$ s.t. $\forall 𝑛\ge {𝑁}_2$, $\left|{𝑦}_{𝑛}-{𝐿}_2\right|<\frac{𝜀}{2}$.

So $\forall 𝑛\ge max\left({𝑁}_1,{𝑁}_2\right)$,

Proof. Suppose for the sake of contradiction that $𝐿>𝑀$ and take $𝜀=\frac{𝐿-𝑀}{2}$.

As ${𝑎}_{𝑛}\to 𝐿$, $\exists {𝑁}_1$ s.t. $\forall 𝑛\ge {𝑁}_1$, ${𝑎}_{𝑛}>𝐿-𝜀=\frac{𝐿+𝑀}{2}$. Similarly, $\exists {𝑁}_2$ s.t. $\forall 𝑛\ge {𝑁}_2$, ${𝑏}_{𝑛}<𝑀+𝜀=\frac{𝐿+𝑀}{2}$.

Then $\forall 𝑛\ge max\left({𝑁}_1,{𝑁}_2\right)$, ${𝑏}_{𝑛}<\frac{𝐿+𝑀}{2}<{𝑎}_{𝑛}$, which is a contradiction.⁠$\square$

Proof. Let $𝜀>0$. Then since ${𝑐}_{𝑛}\to 𝐿$, $\exists {𝑁}_1$ s.t. $\forall 𝑛\ge {𝑁}_1$, ${𝑐}_{𝑛}<𝐿+𝜀$. Similarly, $\exists {𝑁}_2$ s.t. $\forall 𝑛\ge {𝑁}_2$, ${𝑎}_{𝑛}>𝐿-𝜀$.

So $\forall 𝑛\ge max\left({𝑁}_1,{𝑁}_2\right)$, $𝐿-𝜀<{𝑎}_{𝑛}\le {𝑏}_{𝑛}\le {𝑐}_{𝑛}<𝐿+𝜀$. hence ${𝑏}_{𝑛}\in \left(𝐿-𝜀,𝐿+𝜀\right)$, so ${𝑏}_{𝑛}\to 𝐿$.⁠$\square$

Proof.

(sums)

Note that

$$\begin{array}{cc}\left|\left({𝑎}_{𝑛}+{𝑏}_{𝑛}\right)-\left(𝐿+𝑀\right)\right|& =\left|\left({𝑎}_{𝑛}-𝐿\right)+\left({𝑏}_{𝑛}-𝑀\right)\right|\\ & \le \left|{𝑎}_{𝑛}-𝐿\right|+\left|{𝑏}_{𝑛}-𝑀\right|\text{by the triangle inequality}.\end{array}$$

Let $𝜀>0$, then since ${𝑎}_{𝑛}\to 𝐿$,

$$\exists {𝑁}_1\text{s.t.}\forall 𝑛\ge {𝑛}_1,\left|{𝑎}_{𝑛}-𝐿\right|<\frac{𝜀}{2}.$$

Similarly,

$$\exists {𝑁}_2\text{s.t.}\forall 𝑛\ge {𝑁}_2,\left|{𝑏}_{𝑛}-𝑀\right|<\frac{𝜀}{2},$$

so

$$\forall 𝑛\ge max\left({𝑁}_1,{𝑁}_2\right),\left|\left({𝑎}_{𝑛}+{𝑏}_{𝑛}\right)-\left(𝐿+𝑀\right)\right|<𝜀.$$

(products)

Note that

$$\begin{array}{cc}\left|{𝑎}_{𝑛}{𝑏}_{𝑛}-𝐿𝑀\right|& =\left|\left({𝑎}_{𝑛}-𝐿\right){𝑏}_{𝑛}+𝐿\left({𝑏}_{𝑛}-𝑀\right)\right|\\ & \le \left|{𝑏}_{𝑛}\right|\left|{𝑎}_{𝑛}-𝐿\right|+\left|𝐿\right|\left|{𝑏}_{𝑛}-𝑀\right|.\end{array}$$

Let $𝜀>0$. Since ${𝑏}_{𝑛}\to 𝑀$, $\left({𝑏}_{𝑛}\right)$ is bounded, $\left|{𝑏}_{𝑛}\right|<𝐾$ for some $𝐾\in ℝ$.

Since ${𝑎}_{𝑛}\to 𝐿$,

$$\exists {𝑁}_1\in \mathbb{N}\text{s.t.}\forall 𝑛\ge {𝑁}_1,\left|{𝑎}_{𝑛}-𝐿\right|<\frac{𝜀}{2𝑘}.$$

Similarly,

$$\exists {𝑁}_2\in \mathbb{N}\text{s.t.}\forall 𝑛\ge {𝑁}_2,\left|{𝑏}_{𝑛}-𝑀\right|<\frac{𝜀}{2\left(\left|𝐿\right|+1\right)}.$$

Then $\forall 𝑛\ge max\left({𝑁}_1,{𝑁}_2\right)$,

$$\left|{𝑎}_{𝑛}{𝑏}_{𝑛}-𝐿𝑀\right|<𝐾\frac{𝜀}{2𝐾}+\left|𝐿\right|\frac{𝜀}{2\left(\left|𝐿\right|+1\right)}<𝜀.$$

(quotients)

STP that if ${𝑏}_{𝑛}\to 𝑀$, ${𝑏}_{𝑛}\ne 0\ne 𝑀$, then $\frac{1}{{𝑏}_{𝑛}}\to \frac{1}{𝑀}$.

Since ${𝑏}_{𝑛}\to 𝑀\ne 0$, by taking $𝜀=\frac{\left|𝑀\right|}{2}$, there is some ${𝑁}_1\in \mathbb{N}$ s.t.

$$\forall 𝑛\ge {𝑁}_1,\left|{𝑏}_{𝑛}\right|>\frac{\left|𝑀\right|}{2}.$$

So for $𝑛\ge {𝑁}_1$, we have

$$\begin{array}{cc}\left|\frac{1}{{𝑏}_{𝑛}}-\frac{1}{𝑀}\right|& =\frac{\left|{𝑏}_{𝑛}-𝑀\right|}{\left|{𝑏}_{𝑛}\right|\left|𝑀\right|}\\ & \le \frac{\left|{𝑏}_{𝑛}-𝑀\right|}{\frac{\left|𝑀\right|}{2}\left|𝑀\right|}.\end{array}$$

Let $𝜀>0$, then as ${𝑏}_{𝑛}\to 𝑀$ there is some ${𝑁}_2\ge {𝑁}_1$ s.t.

$$\forall 𝑛\ge {𝑁}_2,\left|{𝑏}_{𝑛}-𝑀\right|<\frac{2𝜀}{{\left|𝑀\right|}^2}$$

and

$$\left|\frac{1}{{𝑏}_{𝑛}}-\frac{1}{𝑀}\right|<𝜀.$$

(modulus)

This follows as $\left|\left|{𝑎}_{𝑛}\right|-\left|𝐿\right|\right|\le \left|{𝑎}_{𝑛}-𝐿\right|$.

⁠$\square$

Proof. Since $\left\{{𝑎}_{𝑛}|𝑛\in \mathbb{N}\right\}$ is nonempty and bounded above, a supremum $𝐿=sup{𝑎}_{𝑛}$ exists.

Let $𝜀>0$. By the approximation property, $\exists 𝑁$ s.t.

and further since $\left({𝑎}_{𝑛}\right)$ is increasing, for all $𝑛\ge 𝑁$,

Hence ${𝑎}_{𝑛}\to 𝐿$.⁠$\square$

Proof. Note that since $\left[{𝑎}_{𝑛+1},{𝑏}_{𝑛+1}\right]\subseteq \left[{𝑎}_{𝑛},{𝑏}_{𝑛}\right]$, then $\left({𝑎}_{𝑛}\right)$ is increasing and $\left({𝑏}_{𝑛}\right)$ is decreasing.

As ${𝑎}_{𝑛}<{𝑏}_1$ for all $𝑛$, and ${𝑏}_{𝑛}>{𝑎}_1$ for all $𝑛$, then both sequences converge. Set $𝛼=lim{𝑎}_{𝑛},𝛽=lim{𝑏}_{𝑛}$.

Note that ${𝑎}_{𝑛}\le 𝛼\le 𝛽\le {𝑏}_{𝑛}$, hence $\left[𝛼,𝛽\right]\subseteq {𝐼}_{𝑛}$ for all $𝑛$, and $\left[𝛼,𝛽\right]\subseteq \bigcap {𝐼}_{𝑛}$.

If ${𝑏}_{𝑛}-{𝑎}_{𝑛}\to 0$ then $𝛼=𝛽$. In this case, if $𝑥\in {𝐼}_{𝑛}$ for all $𝑛$, then ${𝑎}_{𝑛}\le 𝑥\le {𝑏}_{𝑛}$ for all $𝑛$, so $𝑥=𝛼=𝛽$ by sandwiching, and $\bigcap {𝐼}_{𝑛}=\left\{𝛼\right\}$.⁠$\square$