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

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- Apr 15th 2009, 07:09 PMmonkey.brainshow 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 AMMoo
Hello,

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