Example of a non-hausdorff space

Question

For instance, the simplest example of a non-hausdorff topological space is the the pair $(X, \tau_X)$ where $\tau_X = \{ X, \varnothing \} $. But this is boring. Can someone help me find more interesting examples?

Answer 1

Perhaps the easiest to visualise is the line with two origins which is just given by gluing two real lines together, point-for-point except for the origins of both lines. I like to think of the line with two origins as taking the real line and then 'puncturing' it at the origin but with a hole that's 'too small to see' - this is of course hand-wavey.

More precisely we define $l$ to be the space $$l=(\mathbb{R}_1\sqcup\mathbb{R}_2)/{\sim}$$ where $(x,1)\sim (x,2)$ for all $x\neq 0$.

The reason $l$ is not Hausdorff is because open sets around either of the origins, which we'll now just call $0_1$ and $0_2$, contains arbitrarily small closed intervals of the type $(0,\epsilon)$. In the broad topological sense of convergence of sequences, we can say that a sequence of points in $l\setminus\{0_1,0_2\}$ converges to $0_1$ if and only if it also converges to $0_2$. This phenomena can not occur in Hausdorff spaces (sequence converging to a set with more than a single element).

This idea of gluing Hausdorff spaces together everywhere except for some closed subset is a good way of creating non-Hausdorff spaces that in some sense 'feel familiar' - we can almost visualise them.

Answer 2

There are lots of examples that you can come up with; see, say, Ian's in the comments.

For the sake of not getting bogged down in the details, I'm going to be a little lax in my descriptions here. One important non-Hausdorff topology that comes up very naturally is the Zariski topology in algebraic geometry. It's easiest to define in terms of closed sets (we can say a set is open iff its complement is closed, so this is perfectly fine). Let's start with, say , $\Bbb R^2$. An algebraic set in $\Bbb R^2$ is a set of points that satisfy some set of polynomial equations (where the polynomials have coefficients in $\Bbb R$). Any given point is closed, as we can just pick the polynomial set to be $\{x-a=0,y-b=0\}$; the only point in the zero set of both equations is $(a,b)$. Of course, curves defined by polynomials are closed sets: say, $y^2-x^3-5x=0$. We define a set to be closed in the Zariski topology if there's some finite set of polynomials such that the zero set of those polynomials is exactly our set.

This topology is not Hausdorff (time for a handwavy proof; skip if you'd like); if we picked two points, we could surely find open sets containing one but not the other. However, are open sets (in the Zariski topology on $\Bbb R^2$) are the complements of finite unions of curves. This means that the vast majority of $\mathbb{R}^2$ is in our open set; in terms of measure theory, all of our (not-whole-space) closed sets have measure $0$ (in the Lebesgue measure on $\Bbb R^2$). Then we can't pick two disjoint sets that each contain exactly one of our points: the intersection of these is also an open set, but clearly not the empty set, since our open sets mutually covered the vast majority of the plane.

Answer 3

A pseudo metric space, i.e. a set with a metric where the distance between two distinct points can also be 0, need not be Haussdorf. For instance, consider the real numbers with an infinitesimal positive element ε, such that there's nothing between ε and 0. Then any open ball (in the pseudo metric) containing 0 will also contain ε, as d(0, ε) = 0, and hence the Haussdorf condition fails. Any sequence converging to ε will also converge to 0.

Answer 4

In order to define lower semi-continuity, it is possible to introduce the following non-Hausdorff (but Kolmogorov) topology: $$\mathcal{T} = \{ [a,+ \infty) \mid a \in \mathbb{R} \} \cup \{ \emptyset \}.$$ Thus, a function $f : \mathbb{R} \to \mathbb{R}$ is lower semi-continuous iff it is continuous with respect to $\mathcal{T}$.

Answer 5

If you're willing to relax $T_1$ (which you might not be OK with), then you can construct a non-Hausdorff topology out of two points:

$X = \{a,b\}$

$\tau_X = \{\{a\},\{a,b\}, X, \emptyset \}$,

which is also a non-discrete topology on a finite set. In general, a non-discrete topology on a finite set should generate a non-Hausdorff space, and any topology on a finite set that is neither discrete nor trivial will require you relax $T_1$.

Answer 6

Another topological space, which is not Hausdorff (unfortunately not T1, either), is the open interval $(0,1)$ with the nested interval topology.

Similarly to the cofinite topology (it's T1 and verifiable almost analogously) the cocountable Topology $$ \mathcal{T} := \{ U \subset X: U = \emptyset \text{ or } U^C \text{ is countable}\} $$ of a infinite set $X$, which is also not Hausdorff.

Answer 7

What about the cofinite topology on $\Bbb{R}$? That is define a topology $\tau$ on $\Bbb{R}$ as follows

$$\tau:=\{U \subset \Bbb{R} \text{is open}: \Bbb{R} \setminus U \text{is finite}\}$$

One can easily prove this space is non Hausdorff. Consider $x \in U,y \in V \in \tau$ with $U,V$ disjoint. Then

$$\Bbb{R} \setminus U = \{p_1,...,p_m\}$$ $$\Bbb{R} \setminus V = \{q_1,...,q_n\}$$

As $\Bbb{R}$ is infinite, there exists $z \in \Bbb{R}$ such that

$$z \neq p_i, q_j \space \text{for $1\leq i \leq m$,$1 \leq j \leq n$}$$

This forces $z \in U \cap V$ and they are not disjoint.