Josia Pietsch c986475c77
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gist for lectures 1-4
2024-01-23 21:52:45 +01:00

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A countable product of separable spaces $(X_n)_{n \in \N}$ is separable.
Choose a countable dense subset $D_n \subseteq X_n$
Fix some point $(a_1,a_2,\ldots) \in \prod_n X_n$
and consider $\bigcup_{i \in \N} \prod_{n \le i} D_n \times \prod_{n > i} \{a_n\}$.
\item Let $X$ be a topological space.
Then $X$ \nth{2} countable $\implies$ X separable.
\item If $X$ is a metric space and separable,
then $X$ is \nth{2} countable.
For the first point, choose some point from every basic open set.
For the second point consider balls of rational radius
around the points of a countable dense subset.
A topological space is \vocab{Lindelöf}
iff every open cover has a countable subcover.
Let $X$ be a metric space.
If $X$ is Lindelöf,
then it is \nth{2} countable.
For all $q \in \Q$
consider the cover $B_q(x), x \in X$
and choose a countable subcover.
The union of these subcovers is
a countable base.
Let $X$ be a topological space.
If $X$ is \nth{2} countable,
then it is Lindelöff.
Let $A_0, A_1,\ldots$
be a countable base.
Let $\{U_i\}_{i \in I}$
be a cover.
Consider $J \coloneqq \{j : \exists i \in I.~A_j \in U_i\}$.
For every $j \in J$ choose a $U_i$ such that
$A_j \subseteq U_j$.
Let $I' \subseteq I$ be the subset of chosen indices.
Then $\{U_i\}_{i \in I'}$ is a countable subcover.
For metric spaces the notions
of being \nth{2} countable, separable
and Lindelöf coincide.
In arbitrary topological spaces,
Lindelöf is the weakest of these notions.
A metric space $X$ is \vocab{totally bounded}
iff for every $\epsilon > 0$ there exists
a finite set of points $x_1,\ldots,x_n$
such that $X = \bigcup_{i=1}^n B_{\epsilon}(x_i)$.