Real Analysis - Combinations of Continuous Functions #2

• Nov 21st 2008, 05:43 PM
ajj86
Real Analysis - Combinations of Continuous Functions #2
Assume h: R --> R is continuous on R and let K = {x: h(x) = 0}. Show that K is a closed set.

Aside: (In the past we have used K to denote a compact set. I'm not sure if it is denoting one here or not. Would they have to specify in the beginning of the problem that K is compact?)
• Nov 22nd 2008, 12:14 AM
Moo
Quote:

Originally Posted by ajj86
Assume h: R --> R is continuous on R and let K = {x: h(x) = 0}. Show that K is a closed set.

It is a closed set because it's the reciprocal image of a closed set by a cont. application.

Proof :
Let's show that the reciprocal image by a continuous application of an open set is open.
Let $\displaystyle x \in f^{-1}(O)=\{x \in R ~:~ f(x) \in O\}$, where O is an open set. Because it's open, it's a neighbourhood of any of its point. Hence it is a neighbourhood of f(x).
Because the function is continuous, $\displaystyle f^{-1}(O)$ will be a neighbourhood of x.
Hence $\displaystyle \forall x \in f^{-1}(O)$, $\displaystyle f^{-1}(O)$ is a neighbourhood of x. So it's a neighbourhood of all its points. So it's an open set.

Now use Hausdorff's formula : $\displaystyle f^{-1}(^cO)=^c f^{-1}(O)$
$\displaystyle ^cO$ is a closed set.
$\displaystyle f^{-1}(O)$ is open, and therefore $\displaystyle ^c f^{-1}(O)$ is a closed set.

We're done.

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$\displaystyle K=h^{-1}(\{0\})$
and in the usual topology over R, {0} is a closed set.

Quote:

Aside: (In the past we have used K to denote a compact set. I'm not sure if it is denoting one here or not. Would they have to specify in the beginning of the problem that K is compact?)
No, but they could ask you to show it is a compact.