In principle a sequence in a nonHausdorff 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?
In principle a sequence in a nonHausdorff 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? 


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 $1/n \to b$. 


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


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 $b$. 


Here are two relevant facts: 1) In a Hausdorff space, a sequence converges to at most one point. 2) A firstcountable 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 firstcountable. See p. 6 for an example showing that 2) does not hold with the hypothesis of firstcountability 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 firstcountable and not Hausdorff, so there must be nonuniquely 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 quasiorder. 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 antisymmetric, i.e., is a partial ordering. Thus in a nonKolmogorov 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 nonKolmogorov space is a pseudometric space which is not a metric space (a pseudometric 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 seminorm which is not a norm always gives such examples. 


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 xaxis y=0. (This topology is not $T_1$.) The constant sequence such that every point is ($\pi$,0) converges to every point on the xaxis. 2) If $\alpha$ and $\beta$ are algebraically independent transcendentals, then the constant sequence {($\alpha$,$\beta$)} converges to every point. Another natural nonHausdorff 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 Ushaped leaves, y=$1/(1x^2)+C$ where 1<x<1. Then a sequence of leaves with $C$ > $\infty$ converges both to the leaf x=1 and the leaf x=+1. 


An easy, nonsilly 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 realvalued integrable functions on $[0,1]$ with the pseudonorm $\f\ = \int_0^1 f$. 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 psuedonorm 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 pseudonorm 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!) 


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, $(x_n) \longrightarrow x$. 


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 nonempty sets have to intersect. 


I can't write a comment, therefore I write an answer. Here http://math.stackexchange.com/questions/1102001/pushoutoftopologicalhausdorffspacesisnothausdorff you can see examples, where the pushout of topological Hausdorffspaces is not Hausdorff. Furthermore this is a way to construct nonHausdorff spaces, construct a suitable pushout.. Regards 

