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\bf{Question}

\quad Decide if these subsets of $\br^2$ are open, closed or
neither:
\medskip

\begin{tabular}{llll}(i)&$\{(x_1,x_2):|x_1|\le3\}$ & \hspace{1cm}(iii)&$\{(x_1,x_2):x_1=0,
x_2<0\}$\\ (ii) &$\{(x_1,x_2):|x_1|<3, x_2>5\}$ & \hspace{1cm}(iv)
&$\{(x_1,x_2):x_1^2+x_1x_2^2>1\}.$
\end{tabular}



\bf{Answer}

\begin{description}
\item{(i)}
Closed: any point $x$ in the complement has $|x_1|>3$, and then
any $x'$ sufficiently close to $x$ (e.g. within $\delta$ of $x$,
where $\delta = |x_1| -3$) has $|x_1|>3$, so the complement is an
open set.

\item{(ii)}
Open: any $x'$ with $|x-x'|<\delta$ will lie in this set if $x$
does and $\delta>0$ is chosen sufficiently small (depending on
$x$). [How small?]

\item{(iii)}
Neither: no point in this set has a $\delta$-nhd contained in the
set. Moreover, $(0,0)$ is the complement but no nhd of $(0,0)$ is.

\item{(iv)}
Open: no need to plot contours etc - just observe that (roughly)
if $x$ doesn't change much then nor does $x_1^2+x_1x_2^2\cdots$.

More precisely, let $y=x+h$: then
$$(y_1^2+y_1y_2^2)-(x_1^2+x_1x_2^2) = 2x_1h_1+h_1^2+
x_1(2x_2h_2+h_2^2) +h_1(x_2^2+2x_2h_2+h_2^2)$$ and the RHS can be
made as small as we please (given $x$) by taking $h$ small enough.
Hence if $x_1^2+x_1x_2^2>1$ then also $y_1^2+y_1y_2^2>1$ for all
$y$ with $y \in B_{\delta}(x)$, some $\delta >0$.

[Remark: This example simply reflects the fact that every
polynomial is continuous.]
\end{description}


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