Exercise 2.2

Question

Let $f$ be an arbitrary complex function on $R^1$, and define
    \begin{align*}
      \varphi(x,\delta) &= \sup\left\{ \lvert f(s) - f(t) vert : s,t \in (x -
      \delta, x + \delta) ight\},\
      \varphi(x) &= \inf\left\{ \varphi(x,\delta) : \delta > 0 ight\}.
    \end{align*}
    Prove that $\varphi$ is upper semicontinuous, that $f$ is continuous at a
    point $x$ if and only if $\varphi(x) = 0$, and hence that the set of points
    of continuity of an arbitrary complex functions is a $G_\delta$.
    \par Formulate and prove an analogous statement for general topological
    spaces in place of $R^1$.

Amit Awasthi · Accepted Answer

\begin{proof} We formulate and prove the statement for general topological spaces $X$, namely that defining for each open set $V \subset X$ and each $x \in X$, \begin{align*} \varphi_V(x) &= \begin{dcases*} \sup_{s,t \in V} \lvert f(s) - f(t) vert & if $x \in V$\ \infty & otherwise \end{dcases*}\ \varphi(x) &= \inf_{V \subset X}\left\{ \varphi_V(x) ight\} \end{align*} that $\varphi(x)$ is upper semicontinuous, and that $f$ is continuous at $x$ if and only if $\varphi(x) = 0$, and the set of points where $f$ is continuous is a $G_\delta$. Note this is equivalent to the definition given for $\mathbf{R}^1$ as our topological space since the intervals $(x - \delta, x + \delta)$ form a basis for the topology on $\mathbf{R}^1$. \par Each $\varphi_V(x)$ is upper semicontinuous since \begin{equation*} \varphi_V^{-1}([-\infty,\alpha)) = \begin{dcases*} V & if $\lvert f(s) - f(t) vert < \alpha$ $\exists s,t\in V$\ \emptyset & otherwise \end{dcases*} \end{equation*} and $\varphi(x)$ is upper semicontinuous since it is the infimum of a family of upper semicontinuous functions. \par We now claim $f$ is continuous at $x$ if and only if $\varphi(x) = 0$. $f$ is continuous at $x$ if and only if for all $\epsilon > 0$, there exists an open neighborhood $V i x$ such that $\lvert f(s) - f(t) vert < \epsilon$ for all $s,t \in V$, but this is true if and only if there exists and open neighborhood $V i x$ such that $\sup_{s,t \in V} \lvert f(s) - f(t) vert < \epsilon$. This in turn is the same as saying $\varphi(x) < \epsilon$ for all $\epsilon > 0$, i.e., $\varphi(x) = 0$. The claim about the set where $f$ is continuous is a $G_\delta$ follows since $f$ is continuous on \begin{equation*} \varphi^{-1}(0) = \bigcap_{n=1}^\infty \varphi^{-1}([-\infty,1/n)) \end{equation*} and each $\varphi^{-1}([-\infty,1/n))$ is open since $\varphi$ is upper semicontinuous. \end{proof}

Exercise 2.2

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Exercise 2.2

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