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HW:  Can either disprove $f: A \rightarrow B$, $f$ bijective and continuous, $A, B \subset {\cal R}$ implies $f$ is a homeomorphism or prove assuming $A$ is open in ${\cal R}$.

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Note $(0, 1) \cup [3, 4] $ has the least upper bound property.  The order topology on this set is 
NOT the
same as the subspace topology.


Note $(0, 1) \cup (3, 4] $  does not have the least upper bound property.  The order topology on this set is NOT the same as the subspace topology.  (0, 1) is not open in the order topology.

Thm 27.1:  Let $X$ by a simply ordered set having the least upper bound property.  In the order topology, each closed interval in $X$ is compact.

Suppose $[a, b] \subset \cup U_\alpha$.  

Let $C =  \{ y \in (a, b] ~|~  [a, y] $ can be covered by a finite number of $U_\alpha \} \cup \{a\}$.

$a \in C$ implies $C \not= \emptyset$,  $C$ is bounded above by $b$.  Hence $sup C$ exists in $X$.



Let $c = sup C$.

Claim:  

\eject


Claim:  $c \in C$.  $c \in [a, b] \subset \cup U_\alpha$.   Hence there exists $\alpha_0$ such that $c \in U_{\alpha_0}$. 

$U_{\alpha_0}$ open implies there exists an $(x, c] \subset U_{\alpha_0}$ for some $x \in [a, b]$.


\eject

Cor:  Every closed interval in ${\cal R}$ is compact.  

Thm 27.3:  A subspace of $R^n$ is compact iff it is closed and bounded in the Euclidean metric or the square metric.
\v
Idea of proof:   $(=>)$  compact Hausdorff implies closed.
\v
If $A$ not bounded, look at $A \subset \cup_{n=1}^\infty B({\bf 0}, n)$  
\v
Idea of proof:   $(=>)$  If $A$ closed and bounded \break
$A \subset B({\bf 0}, r) \subset \Pi [-r, r]$  

Note: a set which is bounded in one metric can be unbounded in a different metric even when both metrics generate the same topology. 

Thm 27.4 (Extreme value thm).  \hfil \break
$f^{cont}: (X, ~compact) \rightarrow (Y, ~ordered)$ implies there exists $c, d \in X$ such that $f(c) \leq f(x) \leq f(d)$ for all $x \in X$.

\v
Idea of proof:  If $f(X)$ has no largest element, then $f(X) \subset \cup_{y \in f(X)} (-\infty, y)$ 


Defn:  If $A$ nonempty subset of metric space $X$ and $x \in X$, then the distance from $x$ to $A$ is

\centerline{$d(x, A) = inf\{d(x, a) ~|~ a \in A\}$}
\v
Note $d_A: X \rightarrow [0, \infty)$, $d_A(x) = d(x, A)$ is a continuous function.
\v
Idea of proof:  Show that $d(x, A) – d(y, A) \leq d(x, y)$


Defn:  The diameter of $A$ = $diam(A) =$ 

\centerline{$sup\{d(a_1, a_2) ~|~ a_1, a_2 \in A\}$}

Lemma 27.5 (Lebesgue number lemma) \hfil \break
 Let $X \subset \cup U_\alpha$.  If $X$ is compact, there is a $\delta > 0$ such that if $diam(C) <
\delta$, then there exists $\alpha_0$ such that $C \subset U_{\alpha_0}$ \v Defn:  $f: (X, d_x)
\rightarrow (Y, d_y)$ is uniformly continuous if for all $\epsilon > 0$, there exists a $\delta > 0$
such that $d_X(x_1, x_2) < \delta$ implies $d_Y(f(x_1), f(x_2)) < \epsilon$. \v Thm 27.6 (Uniform
continuity thm)  \hfil \break $f^{cont}: (X, ~compact ~ metric) \rightarrow (Y, ~metric)$ implies $f$
uniformly continuous. \v Idea of proof:  $X \subset \cup_{y \in Y} f^{-1}(B_{d_Y}(y, {\epsilon \over
2})$.  Let \break
$\delta$ = Lebesgue number of this cover.

Defn:  Given a topological space $X$, $x $ is an isolated point of $X$ is $\{x\}$ is open in $X$.

Thm 27.7:  If $X$ is a nonempty compact Hausdorff space with no isolated points, then $X$ is uncountable.

Step 1:  Take a nonempty open set $U$ and take $x \in X$ (note $x$ may or may not be in $U$).

Use Hausdorff to find nonempty open set $V$ such that $V \subset U$ and $x \not\in \overline{V}$.

Step 2:  Suppose $f: {\cal N} \rightarrow X$, $f(x) = x_n$.  Show $f$ is not surjective (i.e., need to find a point not in the image.  Which definition of compact gives us a point?).
 
\end





$F: N \rightarrow \{{1 \over n}\} \cup \{0\}$

F(1) = 0.  $F(n) = {1 \over n-1}$ for $n > 1$

$F^{-1}: \{{1 \over n}\} \cup \{0\} \rightarrow N$
  
F(1) = 0 which is not open

F: A to B
G: B to A
Take b \in B Take $r > 0$

Suppose $A$ open  there exists $R< r$ such that $B(g(b), R) \in A$

F: [g(b) – R/2, g(b) + R/2] to B’ = [f(c), f(e)]  is a homeomorphism.

G:B’ to [g(b) – R/2, g(b) + R/2]

Thus there exists a d > 0 such that x \in (b – d, b + d) \