\tutorial{01}{2023-10-17}{}


\begin{fact}
    A countable product of separable spaces $(X_n)_{n \in \N}$ is separable.
\end{fact}
\begin{proof}
    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 $\bigcup_{i \in \N} \prod_{n \le i} D_n \times \prod_{n > i} \{a_n\}$.
\end{proof}

\begin{fact}
    \begin{itemize}
        \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.
    \end{itemize}
\end{fact}
\begin{proof}
    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.
\end{proof}

\begin{definition}
    A topological space is \vocab{Lindelöf}
    iff every open cover has a countable subcover.
\end{definition}


\begin{fact}
    Let $X$ be a metric space.
    If $X$ is Lindelöf,
    then it is \nth{2} countable.
\end{fact}
\begin{proof}
    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.
\end{proof}
\begin{fact}
    Let $X$ be a topological space.
    If $X$ is \nth{2} countable,
    then it is Lindelöff.
\end{fact}
\begin{proof}
    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.
\end{proof}
\begin{remark}
    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.
\end{remark}

\begin{definition}+
    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 $X = \bigcup_{i=1}^n B_{\epsilon}(x_i)$.
\end{definition}