MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

closed as off-topic by Steven Landsburg, Joseph Van Name, Stefan Kohl, Alex Degtyarev, Dan Petersen Apr 25 '15 at 21:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Steven Landsburg, Joseph Van Name, Stefan Kohl, Alex Degtyarev, Dan Petersen
If this question can be reworded to fit the rules in the help center, please edit the question.

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
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. – Loop Space Dec 18 '09 at 13:29
The question isn't very hard, but I wonder if closing it is an overreaction. – Greg Kuperberg Dec 18 '09 at 15:33
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
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

10 Answers 10

up vote 10 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 $1/n \to b$.

share|cite|improve this answer
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. – Loop Space Dec 18 '09 at 16:35

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 $$\vec{x}_k = (k!,(k!)!,((k!)!)!, \ldots, \text{$k$ with $n$ factorials}).$$

share|cite|improve this answer

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 $b$.

share|cite|improve this answer

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

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

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!)

share|cite|improve this answer

I can't write a comment, therefore I write an answer. Here you can see examples, where the pushout of topological Hausdorff-spaces is not Hausdorff. Furthermore this is a way to construct non-Hausdorff spaces, construct a suitable pushout.. Regards

share|cite|improve this answer

An easy example, in the same vein as Greg's one. Take the real line $\mathbb{R}$ with the 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, $(x_n) \longrightarrow x$.

share|cite|improve this answer
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

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 Hausdorff, two non-empty sets have to intersect.

share|cite|improve this answer

Consider $X = \mathbb{R}$ , and for every $a \in [-\infty , \infty ] $ , $(a,\infty )$ is an open set.

It is a topology because any finite intersection yields an open "ray" of the largest starting point, and any union yields an open "ray" with the minimal starting point, which might be $-\infty $. It is worth noticing that if we'd choose $[a, \infty )$ as our topology base, we'd result in open rays as well (consider $\cup_j [1/j, \infty)$ for example.

Well, now take the sequence $x_j = 2+ 1/j$. We now show it converges $\forall b\leq 2$. For such $b$ and an arbitrary open set $b \in U$, it has the form of $U = (c, \infty ), c < b <2 < x_j , \forall j$, therefore $x_j$ converges to all $b \leq 2$

This examples relays heavily on the fact that every neighborhood of $b$ is also a neighborhood of $2$ , and therefore you can only say of a sequence what it's minimal neighborhood, but not a point of convergence.

share|cite|improve this answer

protected by Gerald Edgar Apr 25 '15 at 21:00

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.