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## T1 spaceIn topology and related branches of mathematics,T and _{1} spacesR are particularly nice kinds of topological spaces.
The T_{0} spaces_{1} and R_{0} properties are examples of separation axioms.
A T
## Definitions
A topological space - Given any two distinct points
*x*and*y*in*X*, each lies in an open set which does not contain the other. In other words, the singleton sets {*x*} and {*y*} are separated unless*x*=*y*. - Given any
*x*in*X*, {*x*} is a closed set. In other words, the fixed ultrafilter at*x*converges only to*x*. - For every point
*x*in*X*and every subset*S*of*X*,*x*is a limit point of*S*if and only if every open neighbourhood of*x*contains infinitely many points of*S*.*Proof*: Suppose singletons are closed in*X*. Let*S*be a subset of*X*and*x*a limit point of*S*. Suppose there is an open neighbourhood*U*of*x*that contains only finitely many points of*S*. Then*U*\\ (*S*\\ {*x*}) is an open neighbourhood of*x*that does not contain any points of*S*other than*x*. (Here is where we use the fact that singletons are closed.) This contradicts the fact that*x*is a limit point of*S*. Thus, every open neighbourhood of*x*contains infinitely many points of*S*. Conversely, suppose there is a point*x*in*X*such that the singleton {*x*} is not closed. Then there is a point*y*≠*x*in the closure of {*x*}. We claim that any open neighbourhood*U*of*y*contains*x*. For suppose not; then the complement of*U*in*X*would be a closed set containing*x*, and the closure of {*x*} would be contained in the complement of*U*. Since*y*is in the closure of {*x*}, this would force*y*not to be in*U*, contradicting the fact that*U*is a neighbourhood of*y*. We have shown that*y*is a limit point of*S*= {*x*}. But it is clear that*X*is a neighbourhood of*y*that does not contain infinitely many points of*S*. This completes the proof.
X is R if and only if either of the following conditions is satisfied:
_{0}- Given any two topologically distinguishable points
*x*and*y*in*X*, each lies in an open set which does not contain the other. In other words, {*x*} and {*y*} are separated unless*x*and*y*are topologically indistinguishable. - Given any
*x*in*X*, the closure of {*x*} owns only the points that*x*is topologically indistinguishable from. In other words, the fixed ultrafilter at*x*converges only to the points that*x*is topologically indistinguishable from.
A space is T
Do not confuse the term "Fréchet topology", which is equivalent to "T ## Examples
The Zariski topology on an algebraic variety is T
For a more concrete example, let's look at the cofinite topology on an infinite set.
Specifically, let - the open set
*O*_{{x}}contains*y*but not*x*, and the open set*O*_{{y}}contains*x*and not*y*; - equivalently, every singleton set {
*x*} is the complement of the open set*O*_{{x}}, so it is a closed set;
_{1} by each of the definitions above.
This space is not T_{2}, because the intersection of any two open sets O_{A} and O_{B} is O_{A∪B}, which is never empty.
Alternatively, the set of even integers is compact but not closed, which would be impossible in a Hausdorff space.
We can modify this example slightly to get an R -
*U*_{A}:= ∪_{x in A}*G*_{x}.
_{0} (and hence not T_{1}), because the points x and x + 1 (for x even) are topologically indistinguishable; but otherwise it is essentially equivalent to the previous example.## Generalisations to other kinds of spaces
The terms "T
As it turns out, uniform spaces, and more generally Cauchy spaces, are always R | |||||||

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