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Let $X$ be a set. We define the following topology on $X$:

$$\tau=\{S\subseteq X:X\setminus S\text{ is finite}\}$$

How do you prove all neighbourhoods are open in this topology.

Thanks

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Do you mean $\{S\subseteq X:X\setminus S\text{ is finite}\}\cup\{\varnothing\}$? And what is $k$? –  Brian M. Scott Apr 18 '13 at 0:46
    
its the c without the line underneath but apart from that yeah. i couldn't make it look right –  andwil Apr 18 '13 at 0:47
    
sorry k is topology it's supposed to be tau but i don't know how to write things on here, what does it use? –  andwil Apr 18 '13 at 0:48
    
It’s basically just $\LaTeX$; see here to get started. –  Brian M. Scott Apr 18 '13 at 0:49
    
is it in latex? –  andwil Apr 18 '13 at 0:50

2 Answers 2

HINT: $\tau$ is the cofinite topology on $X$. Suppose that $x\in X$ and that $N$ is a nbhd of $x$. Then there is some $V\in\tau$ such that $x\in V\subseteq N$. Clearly $V\ne\varnothing$, so $X\setminus V$ is finite. What does that tell you about $X\setminus N$?

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that it is finite? –  andwil Apr 18 '13 at 0:55
    
Do i not use open balls? that seems to be the answer to everything? –  andwil Apr 18 '13 at 0:57
    
@andwil: Yes: $X\setminus N$ must be finite, and therefore by definition $N\in\tau$, i.e., $N$ is open. You can’t use open balls unless you have a metric, and this is not a metric space. –  Brian M. Scott Apr 18 '13 at 0:59

A neighbourhood $U$ of a point $x$ is any set that contains an open set $V$ containing $x$. So $U$ contains $V$, but $V$ is almost all of $X$. Then $U$ itself must contain all of $X$ minus a finite number of points and hence be open.

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