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Can someone provide me with an example of a non-Hausdorff space. I can't seem to conjure one up
Im sorry but I am struggling to understand this.Thus in the indsicrete topology every sequence tends to every point since there is only one open set (apart from the empty set).
I need to sit down.... A topology on set X is exaclty a collection of open sets satisfying certain conditions (equivalently closed sets satisfying dual conditions).Oxymoron said:So open sets in X are determined by the topology on X?
Again, that is the definition, yes.And a sequence [itex]x_n[/itex] converges to a point [itex]x[/itex] if each open neighbourhood of [itex]x[/itex] contains [itex]x_n [/itex] for [itex]n[/itex] sufficiently large.
So does that mean if a sequence exists in the indiscrete topological space, then the elements of the sequence must reside in an open set containing [itex]x_n[/itex].
And the only open set containing any points is [itex]X[/itex] because, from the topology, every other set is not open, or empty.
Fortunately, I'm already sitting down! If X is any set, the definition of "topology on X" is a collection of subsets of X such that:Oxymoron said:I just want to ask. Because we have defined the topology to be indiscrete, that means that the only open sets are the empty set and itself right? So open sets in X are determined by the topology on X?
??? In any topology a point of a sequence is in some open set! It does happen that, in the discrete topology, the only open set is X itself so the entire sequence is in that set!And a sequence [itex]x_n[/itex] converges to a point [itex]x[/itex] if each open neighbourhood of [itex]x[/itex] contains [itex]x_n [/itex] for [itex]n[/itex] sufficiently large. So does that mean if a sequence exists in the indiscrete topological space, then the elements of the sequence must reside in an open set containing [itex]x_n[/itex].
Let {x_{n}} be any sequence in X. Let x be any point of X. To show that {x_{n}} converges to x, we need only observe that the only open set containing x (X itself) also contains every member of {x_{n}}. With the indiscreet topology, every sequence converges to every member of X!And the only open set containing any points is [itex]X[/itex] because, from the topology, every other set is not open, or empty.