# how to show dense in Finite complement topology

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• Apr 15th 2009, 07:09 PM
monkey.brains
how to show dense in Finite complement topology
Let U be any open set in the finite complement topology on http://www.mathhelpforum.com/math-he...a3150bfd-1.gif.

Show that U is dense in http://www.mathhelpforum.com/math-he...a3150bfd-1.gif
• Apr 16th 2009, 10:35 AM
Moo
Hello,
Quote:

Originally Posted by monkey.brains
Let U be any open set in the finite complement topology on http://www.mathhelpforum.com/math-he...a3150bfd-1.gif.

Show that U is dense in http://www.mathhelpforum.com/math-he...a3150bfd-1.gif

U is an open set in this topology. Let's assume that U is not the empty set.
Hence \$\displaystyle \mathbb{R}\backslash U\$ is finite and is a closed set of this topology.
Similarly, it is obvious that any closed set of this topology is finite, except \$\displaystyle \mathbb{R}\$, which is closed and open.

We have \$\displaystyle U \cup \{\mathbb{R}\backslash U\}=\mathbb{R}\$
But \$\displaystyle \mathbb{R}\$ is an infinite set, and \$\displaystyle \mathbb{R}\backslash U\$ is a finite set.
Thus U is an infinite set.

A definition of U being dense in \$\displaystyle \mathbb{R}\$, is that the unique closed set containing U is \$\displaystyle \mathbb{R}\$ itself.

But a closed set in the finite complement topology is, as said before, finite.
Thus there is no possible closed set containing U, except \$\displaystyle \mathbb{R}\$
This proves that U is dense in \$\displaystyle \mathbb{R}\$

Now you have to study the situation where U=\$\displaystyle \emptyset\$. It looks simple, but I'm quite confused... can you try it ? :p