Non-Hausdorff Space Example: Learn Here

[Oxymoron]

Can someone provide me with an example of a non-Hausdorff space. I can't seem to conjure one up

 

[HallsofIvy]

The "indiscrete" space should work: Let X be any set, the topology consist only of the empty set and X itself. I'll leave it to you to show that that is a valid topology and that it is not Hausdorf.

 

[Galileo]

The real line with the 'half-infinite' topology is not Hausdorff. This is the topology with the sets {[a,->) | a is real}.

 

[Oxymoron]

The reason why I wanted an example of a non-Hausdorff space is one of convergence.

If I have sequence in a Hausdorff space, then the limit of this sequence is unique (ie. if it converges to A and B in a Hausdorff space then A = B).

But is the limit unique in a non-Hausdorff space? That is, in a non-Hausdorff space can a sequence have two, non-identical limits?

 

[matt grime]

limits in any non-hausdordff spave are not necessarily unique (though they may be).

It is important to remember waht it means for a seqquence to converge to something in an arbitrary topological space: x_n tends to x if for all open U with x in U then for all n sufficiently large x_n is in U.

Thus in the indsicrete topology every sequence tends to every point since there is only one open set (apart from the empty set).

Other non-hausdorff spaces include the zariski topology on R (or C or any field). A set is closed iff it is the set of zeroes of a polynomial. Thus the closed sets are precisely finite sets of points We can generalize to R^n (or C^n etc) with a bit of care. This is probably the most "naturally occurring" non-hausdorff space since it is the topology on the algebriac versions of Differential Manifolds (Algebraic varieties).

If a space is Hausdroff and x_n were to tend to x and y with x=/=y in its topology then there are disjoint open sets about x and y U and V say, and we know that all x_n for n sufficiently large are in U and V, but they are disjoint, contradiction, so there cannot be two distinct limits.

this is the second time today I've written that arguemtn out, though the first time was for ordinary metric spaces, well, R with the metric topology.

 

[Oxymoron]

Thanks for replying Matt.

Thus in the indsicrete topology every sequence tends to every point since there is only one open set (apart from the empty set).

Im sorry but I am struggling to understand this.

 

[Oxymoron]

If we consider the indiscrete topological space $(X,\tau)$ where

$$\tau = \{\oslash, X\}$$

is the topology on $X$. Then for any sequence $x_n$, every point $x$ in the sequence is in $X$ (obviously), so $x \in X$. Then $x_n \rightarrow x$, ie the sequence converges to every point in $X$. So if $a,b \, \in X$ and $a \neq b$ then regardless, $x_n \rightarrow a$ and $x_n \rightarrow b$.

And this is because the only open set containing $a,b$ is $X$ (well it certainly isn't $\oslash$). But $X$ also contains every point in $x_n$. Therefore, in the indiscrete topological space, every sequence has every point of $X$ as a limit.

How does this sound?

 

[Oxymoron]

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? And a sequence $x_n$ converges to a point $x$ if each open neighbourhood of $x$ contains $x_n$ for $n$ 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 $x_n$. And the only open set containing any points is $X$ because, from the topology, every other set is not open, or empty.

 

[matt grime]

So open sets in X are determined by the topology on X?

I need to sit down... A topology on set X is exactly a collection of open sets satisfying certain conditions (equivalently closed sets satisfying dual conditions).

And a sequence $x_n$ converges to a point $x$ if each open neighbourhood of $x$ contains $x_n$ for $n$ sufficiently large.

Again, that is the definition, yes.

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 $x_n$.

Erm, since if (X,T) is any topological space (X the underlying set and T its topology) X is an element of T, then this statement is trivially true for all topological spaces.

And the only open set containing any points is $X$ because, from the topology, every other set is not open, or empty.

I don't understand what you're getting at. Please, for peace of mind, would you mind telling me what you think a topological space is?

 

[HallsofIvy]

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?

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:
(1) The empty set is in the collection.
(2) The set X itself is in the collection.
(3) The union of any subcollection of sets in the collection is also in the collection.
(4) The intersection of any finite subcollection of sets in the collection is also in the collection.
Yes, the definition of "open set" is "member of the topology" for X.
The "indiscrete" topology for any given set is just {φ, X} which you can easily see satisfies the 4 conditions above. In the indiscrete topology the only open sets are φ and X itself.
(For any set X, the collection of all subsets of X is also a topology for X, called the "discrete" topology. ALL subsets of X are open (and so also closed) in the discrete topology.)

And a sequence $x_n$ converges to a point $x$ if each open neighbourhood of $x$ contains $x_n$ for $n$ 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 $x_n$.

? 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 the only open set containing any points is $X$ because, from the topology, every other set is not open, or empty.

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!

Exercise for the student: Reverse yourself and show that in the discreet topology, the only convergent sequences are the constant sequences!