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In principle a sequence in a non-Hausdorff space can converge to two points simultaneously.

Can anyone give me an explicit example of the above?

Or tell me any method of generating such kinds of examples?

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Let $X$ be a set with more than one element, and put the indiscrete topology on it: only $X$ and $\emptyset$ are open. Then any sequence will converge to any point of $X$, because each point has only one non-empty neighbourhood: the whole space $X$! –  Konrad Swanepoel Dec 18 '09 at 12:56
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I'm voting to close. As asked, the question is not of a suitable level for MO. Although the wikipedia page on Hausdorffness doesn't list any specific non-Hausdorff spaces, it does give enough information to construct one. If you have a specific case in mind, edit the question to be more precise. –  Andrew Stacey Dec 18 '09 at 13:29
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The question isn't very hard, but I wonder if closing it is an overreaction. –  Greg Kuperberg Dec 18 '09 at 15:33
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I don't think this should be closed. It strikes me as a good example of what the FAQ calls a "standard question ... that mathematicians have when they are exploring a new field". This phenomenon comes up in plenty of situations in algebraic geometry, and it can be counter-intuitive to people who only think in separated toplogical spaces. –  David Speyer Dec 18 '09 at 17:12
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Funny; I thought was going to get closed. I don't like answering questions that will likely be closed, so I commented instead. I am not even sure myself why I think this way – maybe because it feels wrong to gather reputation by providing very easy answers? –  Harald Hanche-Olsen Dec 18 '09 at 18:20
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8 Answers

up vote 9 down vote accepted

Let $X = \mathbb{R} \setminus \{0 \} \cup \{ a,b\}$. Hence $X$ is the real line sans the origin with two points $a\neq b$, both not in $\mathbb{R}$, thrown in. The topology is generated by the open intervals in $\mathbb{R} \setminus \{0\}$ along with sets of the form $(u,0)\cup \{a\} \cup (0,v)$ and $(u,0)\cup \{b\} \cup (0,v)$, where $u < 0 < v$. $X$ is not Hausdorff because $a$ and $b$ cannot be separated by disjoint open sets. Every sequence that converges to $a$ also converges to $b$. Eg. $1/n \to a$ and $1/n \to b$.

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Every sequence that converges to $a$ but does not use $a$. –  Greg Kuperberg Dec 18 '09 at 16:16
    
Greg, you should have just used your superpowers and put that it. –  Andrew Stacey Dec 18 '09 at 16:35
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The easiest type of counterexample is a space that is not $T_1$, which means that there exist two points $x$ and $y$ such that every open set that contains $x$, also contains $y$. If that happens, then every sequence of points that converges to $y$, also converges to $x$. The most extreme case, as Konrad points out, is $X$ with the indiscrete topology. Then everything converges to everything. Examples that are not $T_1$ are valid but artificial. Given such a space, you can make a natural $T_1$ quotient using the closures of all of the points (even though these closures may be nested), and then ask the question again.

An indisputably natural example which is also $T_1$ is the Zariski topology on $\mathbb{Q}^n$. In this topology, a set is closed when it is the solution set to a polynomial equation with rational coefficients. This is a poorly behaved topology, but it is widely used, and (in the version that I am using) points are closed. You can still make a sequence that converges to every point. Number the set of available polynomials $p_1, p_2, \ldots$, and then choose each point $\vec{x}_k$ such that $p_j(\vec{x}_k) \ne 0$ when $j < k$. The construction is also possible in the Zariski topology on $\mathbb{R}^n$, but it is trickier because the polynomials now have real coefficients and there are uncountably many. Nonetheless you can let $$\vec{x}_k = (k!,(k!)!,((k!)!)!, \ldots, \text{$k$ with $n$ factorials}).$$

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here is another example, that shows, that the following statement is (surprisingly) not symmetric: Every sequence that converges to $a$ also converges to $b$.

Consider the set two element set $\{a,b\}$ with topology $\{\emptyset,\{b\},\{a,b\}\}$. Then every sequence, that converges to $a$ also converges to $b$ and the sequence, which is constant $b$ converges only to $b$.

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Here are two relevant facts:

1) In a Hausdorff space, a sequence converges to at most one point.

2) A first-countable space in which each sequence converges to at most one point is Hausdorff.

See e.g. pages 4 to 5 of

http://math.uga.edu/~pete/convergence.pdf

for the (easy) proofs of these facts, together with the definition of first-countable. See p. 6 for an example showing that 2) does not hold with the hypothesis of first-countability dropped.

It seems like a worthwhile exercise to use 2) to find spaces that have the property you want. For instance, the cofinite topology on a countably infinite set is first-countable and not Hausdorff, so there must be non-uniquely convergent sequences.

Addendum: Here are some further simple considerations which unify some of the other examples given.

For a topological space $X$, consider the specialization relation: a point $x$ specializes to the point $y$ if $y$ lies in the closure of $\{x\}$. This implies that any sequence which converges to $x$ also converges to $y$. (If in the previous sentence we replace "sequence" by "net", we get a characterization of the specialization relation.) The specialization relation is always reflexive and transitive, so is a quasi-order.

Note that a topological space is T_1, or separated, iff the specalization relation is simply equality. Thus in a space which is not separated, there exist distinct points $x$ and $y$ such that every net which converges to $x$ also converges to $y$. If $X$ is first countable, we may replace "net" by "sequence".

A topological space $X$ satisfies the T_0 separation axiom, or is a Kolmogorov space, if for any distinct points $x,y \in X$, there is an open set containing exactly one of $x$ and $y$. A space is Kolmogorov iff the specalization relation is anti-symmetric, i.e., is a partial ordering. Thus in a non-Kolmogorov space, there exist distinct points $x$ and $y$ such that a net converges to $x$ iff it converges to $Y$. (If $X$ is first countable...)

An example of a first countable non-Kolmogorov space is a pseudo-metric space which is not a metric space (a pseudo-metric is like a metric except $\rho(x,y) = 0 \iff x = y$ is weakened to $\rho(x,x) = 0$). In particular, the topology defined by a semi-norm which is not a norm always gives such examples.

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I think the Zariski topology from a subfield provides even more natural examples. You can define a Zariski topology from Q on $C^n$ so that the closed sets are zero sets of polynomials with rational coefficients. Then

1) Because $\pi$ is transcendental, the closure of ($\pi$,0) in $C^2$ in this topology is the x-axis y=0. (This topology is not $T_1$.) The constant sequence such that every point is ($\pi$,0) converges to every point on the x-axis.

2) If $\alpha$ and $\beta$ are algebraically independent transcendentals, then the constant sequence {($\alpha$,$\beta$)} converges to every point.

Another natural non-Hausdorff space is the quotient topology on leaves of a foliation. Consider the foliation of $R^2$ by vertical lines x=a for a≤-1 or a≥1, and by parallel U-shaped leaves, y=$1/(1-x^2)+C$ where -1<x<1. Then a sequence of leaves with $C$ -> -$\infty$ converges both to the leaf x=-1 and the leaf x=+1.

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Hello, Doug. The "Zariski topology" you describe is what you get on $\mathbb{C}$-points when you pull back the scheme-theoretic Zariski topology on $\mathbb{A}^n_\mathbb{Q}$ along the field inclusion $\mathbb{Q} \to \mathbb{C}$. Since transcendentals map to generic points over Q, you get the density statement. –  S. Carnahan Jan 10 '10 at 21:25
    
Thanks for the clarification. I was going to mention generic points on varieties like $(\pi,\pi^2)$ but then found that generic points had other possible meanings, and that I don't know the current usage. –  Douglas Zare Jan 10 '10 at 22:29
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An easy, non-silly example (that is perhaps more appealing than the Zariski topology to a student at the level of someone asking this question) is simply to consider the space of real-valued integrable functions on $[0,1]$ with the pseudo-norm $\|f\| = \int_0^1 |f|$. The topology generated by the balls is not Hausdorff, an explicit example of a sequence converging to two points is simply the constant sequence $f_n = 0$, which converges both to the constant $0$ function as well as the function $f(x) = 0$ for $x \in [0,1)$, $f(1) = 1$.

While simply considered as a topological space, this really doesn't present any issues, because we may easily quotient to get a Hausdorff space. But while this is trivial from a topological perspective, and we don't lose any information about behavior in the psuedo-norm by quotienting to get a norm, quotienting like that is really quite a violent act as far as pointwise behavior is concerned. We now have to worry about things like sets of measure 0 piling up (on uncountable families) or, likewise, the stark realization that via our a.e. equivalence we improve the behavior of one topology (going from a pseudo-norm to a norm) at the expense of destroying another (from pointwise convergence to a.e. convergence we have abandoned the realm of topology altogether. A.e. convergence does not generally come from a topology!)

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An easy example, in the same vein as Greg's one. Take the real line $\mathbb{R}$ with the finite complement topology, http://en.wikipedia.org/wiki/Finite_complement_topology .

That is, a subset $U \subset \mathbb{R}$ is open if and only if it is the empty set or its complement $\mathbb{R}\backslash U$ is a finite set. Then every sequence $(x_n)$ of points of $\mathbb{R}$ converges to every point $x \in \mathbb{R}$.

To see this, take any open set $U$ containing $x$. Because $\mathbb{R} \backslash U$ has only a finite number of points, an infinite number of points of the sequence $(x_n)$ must be in $U$; i.e., there exists $n_0 \in \mathbb{N}$ such that, for every $n \geq n_0$, $x_n \in U$. Thus, $(x_n) \longrightarrow x$.

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It is exactly the Zariski topology on $\mathbb{R}$, not just in the same vein. –  Greg Kuperberg Dec 18 '09 at 17:11
    
@Greg. Ok. You're right. I forgot to explain this. –  a.r. Dec 18 '09 at 20:07
    
To be picky, (I think) you're requiring your sequences to have infinitely many distinct terms to apply this argument. For example, the constant 0 sequences only converges to 0, not to anything else. –  Jason DeVito Dec 18 '09 at 20:30
    
@Jason. :-) Ok. –  a.r. Dec 19 '09 at 16:29
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Note also that in a T2 space, since you can separate points then the limit will be unique. However that does not mean the converse is true.

We can construct a space in which the limit is unique but the space is not T2. Let the real line have the cocountable topology. Suppose you have a sequence that has 2 limits $x$ and $y$, then consider an open set, call it $U_x$ consisting of the complement of the points which are not x. Then $x\in U_x$ and there must be some $N$ such that $\forall n>N, x_n\in U_x$ that point but $\forall n>N, x_n=x$ because we get $x_n\in U_x\cap\(x_k)=x$, I mean the set of all $x_k$ here. Similarly for $y$ and so now $\forall n>max(N,N')$ we get $x_n=x=y$ which is false since these are two different elements. So the limit is unique.

The topology is not Haussdorff, two non-empty sets have to intersect.

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