Consider the set P := {(x,y) $\displaystyle \epsilon$ $\displaystyle R^2$ : $\displaystyle y^2 = 2x$}. Show that P is closed in $\displaystyle R^2$ but not compact.
Consider the set P := {(x,y) $\displaystyle \epsilon$ $\displaystyle R^2$ : $\displaystyle y^2 = 2x$}. Show that P is closed in $\displaystyle R^2$ but not compact.
thanks in advance!!
Nicole
$\displaystyle P$ is closed because the polynomial $\displaystyle f(x,y)=y^2-2x$ is continuous (everywhere) and $\displaystyle P=f^{-1}(0).$ the reason that $\displaystyle P$ is not compact is that it's clearly not bounded.