how to show dense in Finite complement topology

**monkey.brains** · April 15th 2009, 07:09 PM

Let U be any open set in the finite complement topology on

.

Show that U is dense in

**Moo** · April 16th 2009, 10:35 AM

Hello,

Originally Posted by monkey.brains

Let U be any open set in the finite complement topology on

.

Show that U is dense in

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

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

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

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

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