"Prove that in any topological Hausdorff space all subspaces consisting of one element are closed."

>>7696603
Great job, OP. You disproved topology.

Let x be any element of X.

For every y unequal to x there exists an open neibhoorhood U_y that includes y but not x.

The union of all those U_y is X \ {x} and since it's the union of open sets it must be open too.

Therefore {x} is closed.

>>7696618
Oh great, because that was obviously what I was trying to do.

Got anything useful to add about what is wrong with either my reasoning or the statement?

>>7696603
The discrete topology isn't very Haussdorfy, is it?

>>7696624
With the discrete topology every set is both open and closed. There is no contradiction there.

>>7696623
Thanks

>>7696627
A space X is Hausdorff if for every x, y in X with x not equal to y there exist open neighbourhoods V and W of x, y respectively so that the intersection of V and W is the empty set.

If X is a space endowed with the discrete topology, for every x, y in X (x again unequal to y), {x} and {y} are (open) neighbourhoods of x, y so that their intersection is empty.

Unless Ive done something wrong the discrete topology is kind of Hausdorffy.

>>7696630
Thanks -- I thought the statement implied every subspace should be exclusively closed.

>>7696603
There's nothing stopping a set from being both open and closed, or neither. X itself is open and closed regardless of the topology you give it, as well as the empty set.

To prove the statement just choose a point x in X and use the Hausdorff condition to find open sets disjoint from {x} containing any other point in the space. Union those up and you've got the complement of {x}.

>>7696627
Sure it is, it is even metrizable

Each point in X is a connected component. Imagine a discrete set of points in R^n (such as the integers) if you like.

>>7696603
>stumbling over clopen sets
Babby tier physics student detected. Fuck off, cunt.