Important problems

3

(important!!)

Show $\Q \cup \bdry{\Q} = \R$

Show if $x \in \R$ $x$ is an adherent point of $\Q$.

i.e. there exists a sequence of rational $E$’s $(q^{(n)})^\infty_{n=1}$ in $Q $.

$q^{(n)} \to X_0 $ $d_(q^{(n)}, x_0) \to 0$

Consider

$(x_0 - 1, x_0 + 1) \subseteq \R, B(x_0, 1) \subseteq \R$

$\exists$ rational here $q^{(1)}$

$(x_0 - \frac{ 1 }{ n }, x_0 + \frac{ 1 }{ n })$ has a rational $q^{(n)}$

$d(x_0, q^{(n)}) < \frac{ 1 }{ n }$

$\limu{ n }{ \infty } d(x_0, q^{(n)}) = 0 $

$q^{(n)} \to x_0, x_0 \in \overline{\Q}$

What to take away from it? What is a closure?

Remember

$E$ is closed iff $\overline{ E } = E $ iff $x^{(n)} \in E$ & $(x^{(n)})^{ \infty }_{n=1} \to x_0 \in X$ then $x_0 \in E$.

7

(not in exam!)

Show $X = (a^{(n)})^\infty_{ n =1}: a_n \in \set{ 0, 1 }$

view each entry as vectors

Take a Cauchy sequence in $X$.

$X^{(1)} = (X^{(2)}_1, X^{(2)}_2, …)$

$X^{(1)} = (X^{(1)}_1, X^{(2)}_2, …)$

lower index = entry of the vector upper index = term of sequence

If $(x^{(n)})^\infty_{ n =1} $ is cauchy

$\exists N $ such that if $n, m > N $

then $d(x^{(n), x^{(m)}}) < 2 $ then $x_0 = y_0 $

$\sum_{ j=0 }^{ \infty } \frac{ 1 }{ 2^j } \abs{ x_j - y_j } < 2$

$\frac{ 1 }{ 2^0 } \abs{ x_0 - y_0 } + \sum_{ j=0 }^{ \infty } \abs{ x_0 - y_0 } < 1$

contradiction because:

if $x_0 \neq y_0 $, $d(x, y) \geq 1 $

If n > N

The zeroth entry

$x_0^{(n)} = x_0^{(N+1)} $

0th entry of a possible by $(x_0^{n+1})$

8

$K_1 \subseteq K_2 \subseteq K_3 $ compact sets

$\bigcap_{ n=1 }^ \infty \neq \emptyset$.

We will work in the space $X = K$, ($X$ is compact).jo

$K_n $ compact $\implies$ closed

Notice $X \setminus K_n = O_n$ is open in $X$.

Assume

$\cap K_n = \emptyset $

Take complement

$\bigcup_{ n=1 }^ \infty (X \setminus K_n) = X$

$\bigcup_{ n=1 } ^ \infty O_n = X$ Compact

We can find

$O_1, _2, …, _n_k $ such that

$O_{n_1} \cup O_{n_2} \cup … \cup O_{n_k} \subseteq X$

Notice $X \setminus K_n = O_n $ is open in $X_0$.

$m = \max{ n_1, n_2, …, n__k }$

$O_m \subseteq X $

$K_m = \emptyset$

If $K$ is compact then $K$ is closed.

If $K$ is compact then $K$ is complete.

If $K$ complete \implies $K$ is closed.

If $E \subseteq K $, $K $ complete and $E $ closed, then $E $ is complete.

12

$f: S \to X $ injective. Show $d_s(s_1, s_2) = d_X(f(s_1), f_(s_2)) $ is a metric.

step 0:

Show $0 \leq d_s(s_1, s_2) < \infty $

$f(s_1) \in x_1, f(s_2) \in x_2 $.

$o \leq d_X(x_1, x_2) \leq \infty $

$o \leq d_X(f(s_1), f(s_2)) \leq \infty $

$o \leq d_s(s_1, s_2) \leq \infty $

step 1:

$d_s(s, s) = d_X(f(s), f(s)) = 0$

step 2:

If $s_1 \neq s_2 $, show $d_s(s_1, s_2) > 0$

step 3:

$d_s(s_1, s_2) = d_s(s_2, s_1)$

13

Show if $r_n > 0 $ such that $r_n \to 0 $ with the prop

$d(f(x), f^{(n)}(x)) < r^n $ then $f^{(n)} \to f $ uniformly.

$r_n \to 0 $ if $\forall \epsilon > 0, \exists $ such that $\abs{ r_n -0 } < \epsilon$ if $n > N , r_n < \epsilon$.

$d(f^{(n)}(x), f(x)) < r_n < \epsilon, \forall x \in X$.

Suppose $f^{(n)} \to f $ uniformly, build $r_n \to 0 $

if you want to build a sequence, try $\frac{ 1 }{n} $

14

(!! important)

Recall $d_ \infty (r, g) = \supr_{x \in X} d(f(x), g(x))$

Assume $\limu{ n }{ \infty } d_ \infty(f^{(n)}, f) = 0 $

Let $\epsilon > 0 $, $\exists N $ such that if $n > N $

$\abs{ d_ \infty(f^{(n)}, f) - 0 } < \epsilon $

$d_i \infty(f^{(n)}, f) < \epsilon $

$\supr_{x \in V}{ d(f^{(n)}(x), f(x)) < \epsilon} $

Thus,

$d(f^{(n)}(x), f(x)) < \epsilon, \forall x \in X $

Assume $f^{(n)} \to f $ uniformly show $\forall \epsilon > 0, \exists N$ such that if $n > N $

$d_ \infty(f ^{(n)}, f) < \epsilon$

Let $\epsilon > 0 $

$d(f^{(n)}(x), f(x)) < \epsilon $\forall x \in X$

$\supr_{x \in V}{ d(f^{(n)}(n), f(x))} \leq \epsilon /2 < \epsilon $

15

!!! note If $x < S$, $\forall x \in E \subseteq \R$ $\supr{ E } \leq S$

if $d(f^{(n)}(x), f(x)) < \epsilon $

then $\supr{ d(f^{(n)}(x), f(x))} \leq \epsilon $

$f, g $ are bounded, show $0 \leq d_ \infty(f, g) < \infty $

Notice $0 \leq d(f(x), g(x)), \forall x \in X$, $0 \leq \supr{ x \in V } $

$f $ is bounded, $\exists y_1, r_1 $

$f(x) \subseteq B(y_1, r_1)$

$g \to g(x) \subseteq B(y_2, r_2)$

$d(f(x), g(x)) \leq d(f(x), y_1) + d(g(x), y_2) + d(y_1, y_2) \leq r_1 + r_2 + (y_1, y_2) < \infty$

$\supr_{x \in X}{ d(f(x), g(x))} \leq r_1 + r_2 + d(y_1, y_2) < \infty $

step 1:

$d(f, f) $

$= \supr{ d(f(x), f(x))} $

$= \supr 0 $

$= 0 $

!!!

If $f \neq g $ then $\exists x_0 \in X $ such that $f(x_0) \neq g(x_0) $

$0 < d_Y(f(x_0), g(x_0)) \leq \supr _{x \in X}{ d(f(x), g(x))} $

16

By problem 14, we are done.

17

$f^{(n)}(-1) = (-1^n) $ no convergence

If $x_0 \in (-1, 1) \limu{ n }{ \infty } x_0^{(n)} = 0$

I partially define $f $ $f(x_0) = 0 $

If $x_0 = 1 $

$f^{(n)}(1) = 1^n = 1 $

$\limu{ n }{ \infty } 1 = 1 $

Define $f (1) = 1 $

$ f(x) = \begin{cases} 0 & if x \in (-1, 1) \br 1 & if x = 1 \end{cases}$

Is a pointwise limit of $f^{(n)} $

But $f^{(n)} \to f$ uniformly

$f^{(n)} $ is cont, $f $ is not cont

21

!!!! important

Look at your notes online!

state a version of weistrass M-test

$f^{(n)}: X \to R $(complete) $\to f^{(n)} \in B(X, \R^5) $

$\abs{ f^{(n)}(x)} \leq M_n $(bounded)

$\sum_{ n=0 }^{ \infty } M_n < \infty $

$ f(x) + g(x) $

step 1

If $ g^{(n)} $ is Cauchy is $B(X,Y)$ and $ Y $ is complete then $\exists g $ such that $ g^{(n)} \to g$ in $ d_ \infty $ metric

$B(X, Y)$ is complete if $ Y $ is complete

22

23

24

$\sum_{ n=1 }^{ \infty } f^{(n)} $ converges uniformly

Need is an upper bound

on $f(n) = \frac{ }{ x^2 + y^2 + n^2 }, \frac{ }{ x^2 + y^2 + n^2 } $

$\norm{ f^{(n)}(x, y)} \leq \frac{ 1 }{ x^2 + y^2 + n^2 }, \frac{ 1 }{ x^2 + y^2 + n^2 } $

$\leq \abs{ \frac{ 1 }{ n^2 } + \frac{ 1 }{ n^2 }} $

Show $e^x = \sum_{ n=0 }^{ \infty } \frac{ x^n }{ n! } $ converges uniformly on $[- R, R] $ pointwise convergence

To show $\sum_{ n=0 }^{ \frac{ x^n }{ n! }}$ converges uniformly on $[- R, R] $.

$f^{(n)}(x) = \frac{ x^n }{ n! } $

$\abs{ f^{(n)}(x)} = \abs{ \frac{ x^n }{ n! }} \leq \frac{ R^n }{ n! } = \norm{ f }_ \infty$

$\abs{ f^{(n)}(x)} \leq \frac{ R^n }{ n! } (= M_n) $

$\sum_{ n=0 }^{ \infty } M_n < \infty$ M_n real #?

Ratio test

$\sum_{ n=0 }^{ \infty } \frac{ R^n }{ n! } \to \limu{ n }{ \infty } \frac{ M_n }{ m_{n-1}} < 1$

$\frac{ M_n }{ M_{n-1}} = \frac{ R^n }{ n! } \frac{(n-1)! }{ R^{n-1}} = \frac{ R }{ n }$

$\limu{ n }{ \infty } \frac{ M_n }{ M_{n-1}} = 0 < 1$

Yes the series converges.

By the weierstrass M-test, $\sum_{ n=0 }^{ \infty } \frac{ x^n }{ n! } $ converges uniformly on $[-R, R] $.

Ignore below —————–

Pointwise convergence on $[- \infty, \infty] $.

$\sum_{ n=0 }^{ \infty } \abs{ \frac{ c_n }{ c_{n-1}}} = L$

The radius of convergence is $ R = \begin{cases} 0 &, if L = \infty \br \frac{ 1 }{ L } &, if 0 < L < \infty \br \infty &, if L = 0 \end{cases} $

Ignore above —————–