Monday:
- Review: July 29.
- Homework: Solutions for
homework 5.
- Study: Finish Section 28
- Do: We have an exam coming up Thursday, so just one
problem today called (EP-1). We want to see that the countable
product $[0,1]^{[0,1]}=\prod_{x\in[0,1]}[0,1]$ is compact, this is
an example of a compact space that is not sequentially compact. I
suggest the following. We will accept that every $x\in [0,1]$ has
a binary expansion $x=0.x_1 x_2 x_3 \dots$ so that $$ x =
\sum_{i=1}^\infty x_i\frac1{2^i}.$$ We will also accept this
expansion is unique provided we agree to replace any representation
ending in all $1$s with one ending in all zeros. (Note that, for
example, $0.01111\dots= 0.1$.)
- Recall that elements of $[0,1]^{[0,1]}$ are simply functions
from $[0,1]$ to $[0,1]$. Furthermore, a sequence $(f_n)\subset
[0,1]^{[0,1]}$ converges to $f$ if and only if $f_n(x)\to f(x)$
for all $x\in [0,1]$.
- Let $f_n$ be the function such that $f_n(x)=x_n$ where
$x_n$ is the $n^{\rm th}$ digit of $x$'s binary expansion as
above.
- Suppose that $(f_{n_k})$ is any subsequence of $(f_n)$.
Let $y=0.y_1 y_2 y_3\dots$ be the real number in $[0,1]$ such
that $$y_n=\begin{cases}1 &\text{if $n=n_k$ and $k$ is even,
and} \\ 0 &\text{otherwise.}\end{cases}$$ (For example, if
$n_k=2k$, then $y=0.00010001\dots$.)
- Observe that $f_{n_k}(y)$ does not converge, and conclude
that $[0,1]^{[0,1]}$ is not sequentially compact.
- Due: Wednesday, July 31.
|
Wednesday:
- Solutions: Homework 6.
- Review: July 31.
- Study: Read Section 30.
- Do: EP-2 (Optional -- you don't have to turn this one in).Let
$A$ be a nonempty set and $\mathcal P(A)$ the set of subsets of
$A$. Show that there is no surjection $f:A\to \mathcal P(A)$.
Conclude that $\{0,1\}^\omega=\prod_{n\in \mathbf Z_+}\{0,1\}$ is
uncountable. (Hint: show that $B=\{\,a\in A:a\notin f(a)\,\}$ is
not in the range of $f$.)
- Due: Wednesday, August 7,
|
