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\bf{Question}

\quad The function $f:\br\to\br$ given by
$$f(x)=\cases{1,&$x\ge0$\cr -1,&$x<0$\cr}$$ is {\it not}
continuous at $0$.  Give an example of an open set $V\subset\br$
for which $F^{-1}(V)$ is not open, and a closed set $C\subset\br$
for which $F^{-1}(C)$ is not closed.
\smallskip
Do likewise for the functions $g,h\ $ :
\medskip
$$g(x)=\cases{2x+3,&$x\ge1$\cr -x+4,&$x<1$\cr} \qquad
  h(x)=\cases{\sin({1\over x}),&$x>0$\cr
               0,&$x\le0$\cr}.$$


\bf{Answer}

Let

$V=(0,\infty)$; then $f^{-1}(V)=[0,\infty)$: not open

$C=(-\infty, 0]$; then $f^{-1}(C)=(-\infty,0)$: not closed.

Let

$V=(4,\infty)$; then $g^{-1}(V) = (-\infty,0) \cup [1,\infty)$:
not open

$C=(-\infty, 4]$; then $g^{-1}(C)=[0,1)$: not closed

Let $V=(4,\infty)$; then $g^{-1}(V) = (-\infty, 1) \cup
(1,\infty)$, i.e. $\Re\backslash \{ 1 \}$. Then $h^{-1}(V)$
consists of the positive $x$-axis with all points $\displaystyle
\frac{1}{2n\pi + \frac{\pi}{2}}$, $n \in \textbf{N}$, removed,
together with the negative $x$-axis including $0$. Thus there is
no interval $(-\epsilon, \epsilon)$ $(\epsilon > 0)$ entirely
contained in $h^{-1}(V)$, but $0\in h^{-1}(V)$.

For $C$ take the complement of $V$, namely $\{ 1 \}$. Then all
points $a_n = (2n\pi + \frac{\pi}{2})^{-1}$ lie in $C$, and
$a_n\to 0$ as $n\to\infty$, but $0\notin C$.

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