Monday:
 Study: The previous week's assignments (1123) are due today.
 Do:
 Let $K$ be a compact subset of a metric space $X$ and let
$K \subset U$ be open. Show that there is an open set $V$
such that $K\subset V\subset \overline V\subset U$.
(Consider homework problem #7.)
 Show that $X$ is a Baire space if and only if whenever a
countable union $\bigcup F_n$ of closed sets in $X$ has
interior in $X$ at least one of the sets $F_n$ has interior
in $X$.
 (In this problem, we will assume that if $(X,\rho)$ and
$(Y,\sigma)$ are metric spaces then so is $(X\times
Y,\delta)$ where
$\delta((x,y),(x',y'))=\rho(x,x')+\sigma(y,y')$. You can
also assume that with respect to this product metric,
$(x_n,y_n)\to (x,y)$ if and only if $x_n\to x$ and $y_n\to
y$. In particular, if $(X,\rho)$ and $(Y,\sigma)$ are
complete, so is $(X\times Y,\delta)$.) Let $U$ be a nonempty
open subset of a complete metric space $(X,\rho)$. Show that
$U$ admits a complete metric which is equivalent to that
inherited from $X$. I suggest the following.
 It suffices to find a homeomorphism
$\phi:(U,\rho)\to (Y,\sigma)$ where $(Y,\sigma)$ is
complete.
 Let $A=X\setminus U$ and define $f:U\to \mathbf R$
by $f(x)=\rho(x,A)^{1}$. Then the map $\phi(x)=(x,f(x))$ is
continuous from $(U,\rho)$ to $(X\times\mathbf
R,\delta)$ where $\delta$ is obvious complete product
metric. It suffices to see that that the range of
$\phi$ is closed.

Wednesday:
 Study:
 Do:
 The ruler function is an example of a function
$f:\mathbf R\to \mathbf R$ which continuous at every irrational
and discontinuous at each rational. In this problem, we want
to see that it is impossible to construct a function which is
continous exactly on the rationals. In fact, we are to prove
that if $D$ is a countable dense subset of $\mathbf R$, then there is no
function $f:\mathbf R\to \mathbf R$ such that the set of points
$C$ where $f$ is continuous is equal to $D$. I suggest the following.
 Let $U_n$ be the union of all open sets $U\subset
\mathbf R$ such that $\operatorname{diam}(f(U))<\frac1n$.
Show that $C=\bigcap_n U_n$. (A subset of $\mathbf R$, such
as $C$, which is the countable intersection of open sets is
called a $G_\delta$ subset).
 Show that $D$ can't be a $G_\delta$ subset. (Consider: if
$D=\bigcap W_n$ and $V_d:=\mathbf R\setminus \{d\}$ for each
$d\in D$, then $W_n$ and $V_d$ are dense open subsets of
$\mathbf R$.
 Every vector space $V$ has a basis  that is, a
linearly independent subset $B$ such that every element in
$V$ is a finite linear combination of elements of $B$. The
dimension of $V$, $\operatorname{dim} V$, is the cardinality
of any such basis. (In analysis, such a basis is sometimes
called a Hamel basis to stress that it is a bonifide
vector space basis.) Show that if $V$ is a Banach space,
then its dimension is either finite or uncountable. (Use
problem #25.)
 For Fun Only: The existence of continuous functions that
fail to have a derivative at any point (aka nowhere
differentiable) was greeted with sckepticism when Wierestrass
first proved such things existed. He was forced to produce an
example. (Spivak produces a simpler version of Wierestrass's
example in his Calculus book (see Chapter 23, Theorem 5).)
Using the Baire Category Theorem, we can easily see that the set of
continuous nowhere differentiable functions is dense in
$C[0,1]$. My proof of this is attached
for your amusement.
 Suppose $X$ and $Y$ are Banach spaces with $T\in \mathcal
L(X,Y)$. Suppose that $E$ is a closed proper subspace of
$X$ such that $E\subset \ker T$. Show that there is a unique
operator $\overline{T}\in\mathcal L(X/E,Y)$ such that
$\overline{T}(q(x))=T(x)$ for all $x\in X$ where $q:X\to X/E$ is
the quotient map. Moreover, $\\overline T\=\T\$.
 Suppose that $X$ and $Y$ are Banach spaces, that $D$ is a dense
subspace of $X$ and that $T_0\in\mathcal L(D,Y)$. Show that there
is a unique $T\in \mathcal L(X,Y)$ such that $T(x)=T_0(x)$ for all
$x\in D$. (Let $(x_n)$ and $(y_n)$ be sequences in $D$ converging
to $x\in X$. Show that $(T(x_n))$ and $T(y_n))$ must converge to
the same element of $y$.)

Friday:
 Study: Problems 24 to 43 are due WEDNESDAY. I'll be
posting selected
homework solutions (Last modified December 31, 1969) from time
to time. Keep an eye on the date stamp for new
additions. In particular, I may update the
solutions for the previous week as I see where folks
had trouble when I get around to grading them.
 Do:
 Let $E$ and $X$ be Banach spaces with $E$ finite dimensional.
 Show that every linear map $S:E\to X$ is bounded.
 Show that a linear map $T:X\to E$ is bounded if and
only if $\ker T$ is closed.
 Suppose that $E$ and $M$ are closed subspaces of a Banach
space $X$. If $E$ is finite dimensional, show that $E+M=\{\,
x+y : \text{$x\in E$ and $y \in M$}\,\}$ is closed.
