set closed but not compact

**thaopanda** · September 25th 2009, 04:52 PM

need some help with this...

Consider the set P := {(x,y) $\epsilon$ $R^2$ : $y^2 = 2x$ }. Show that P is closed in $R^2$ but not compact.

thanks in advance!!
Nicole

**NonCommAlg** · September 25th 2009, 08:40 PM

Originally Posted by thaopanda

need some help with this...

Consider the set P := {(x,y) $\epsilon$ $R^2$ : $y^2 = 2x$ }. Show that P is closed in $R^2$ but not compact.

thanks in advance!!
Nicole

$P$ is closed because the polynomial $f(x,y)=y^2-2x$ is continuous (everywhere) and $P=f^{-1}(0).$ the reason that $P$ is not compact is that it's clearly not bounded.

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