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Update exercise
Signed-off-by: Marcello Seri <[email protected]>
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1-manifolds.tex

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\end{enumerate}
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This is where differential geometry comes into play.
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The rest of this chapter will be devoted to the introduction of \emph{smooth manifolds}, which are a class of topological spaces on which it is possible to make sense of the notion of differentiation even though they are not necessarily vector spaces.
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The rest of this chapter will be devoted to the introduction of \emph{smooth manifolds}, which are a class of topological spaces on which it is possible to make sense of the notion of differentiation---even though they are not necessarily vector spaces---and which allows us to reason in a way that will not depend on the way we define coordinates on them.
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We will do this in two stages.
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First we will introduce \emph{topological manifolds}, which are topological spaces that \emph{locally} look like euclidean spaces.
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Then we will endow topological manifolds with a so-called \emph{smooth structure}.
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A topological space $(X, \cT)$ is \emph{Hausdorff} if every two distinct points admit disjoint open neighbourhoods. That is, for every pair $x\neq y$ of points in $X$, there exist open subsets $U_x, U_y\in\cT$ such that $x\in U_x$, $y\in U_y$ and $U_x \cap U_y = \emptyset$.
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\end{definition}
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Topological spaces are extremely general, as such they may have very inconvenient -- someone would say nasty -- properties.
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Topological spaces are extremely general, as such they may have very inconvenient---someone may say nasty---properties.
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You can see this for yourself with the following exercise.
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\begin{exercise}
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Let $X$ be an arbitrary set. Show that $\cT:=\{\emptyset, X\}$ defines a topology on $X$, called the \emph{trivial topology}. Show that on $(X, \cT)$ any sequence in $X$ converges to every point of $X$, and every map from a topological space into $X$ is continuous.
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\begin{itemize}
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\item Let $X$ be an arbitrary set. Show that $\cT:=\{\emptyset, X\}$ defines a topology on $X$, called the \emph{trivial topology}. Show that on $(X, \cT)$ any sequence in $X$ converges to every point of $X$, and every map from a topological space into $X$ is continuous.
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\item Let $X$ be an arbitrary set. Show that $\cT:=\mathcal{P}(X) := \{ A \mid A\subset X \}$, the powerset of $X$, defines a topology on $X$, called the \emph{discrete topology} in which every map $f : X \to Y$ to some other arbitrary topological space $(y, \cU)$ is continuous.
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\end{itemize}
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\end{exercise}
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Hausdorff spaces are still rather general: in particular, any metric space with the metric topology\footnote{Recall that in a metric space $X$ the \emph{metric topology} is defined in the following way: a set $U\subset X$ is called open if for any $x\in U$ there exists $\epsilon>0$ such that $U$ fully contains the ball of radius $\epsilon$ around $x$.} is Hausdorff.

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