Exercise 8.2.9

Question

\begin{enumerate}[label=(\alph*)] 
\item Show that the set $Y =\{f \in C[0,1] : \|f\|_\infty \leq 1 \}$ is closed in $C[0,1]$.
\item Is the set $T = \{f \in C[0,1]: f(0) = 0\}$ open, closed, or neither in $C[0,1]$?
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item $Y^c = \{f \in C[0,1] : \|f\|_\infty > 1\}$. Let $f \in Y^c$ and $\|f\|_\infty = 1 + \epsilon$. Then if $g \in V_{\epsilon/2}(f)$, we have $\|g\|_\infty + \|f-g\|_\infty \geq \|f\|$, implying $\|g\|_\infty \geq 1 + \epsilon / 2$ and so $Y^c$ is open and $Y$ is closed. \item $T$ is not open; the function $f(x) = 0$ is in $T$, but $g_\epsilon(x) = \epsilon/2 \in V_\epsilon(f)$. $T$ is closed; let $f$ be a limit point of $T$. We have that for any $\epsilon > 0$, $\exists h$ so that $\|f-h\|_\infty < \epsilon$, with $h \in T$. Then $\left|f(0)\right| < \epsilon$, which implies $f(0) = 0$ and hence $f\in T$. \end{enumerate}

Exercise 8.2.9

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Exercise 8.2.9

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