# Proof that limit in a Hausdorff space is unique

This proof is likely quite trivial, but I was hoping someone could look it over regardless. There is one particular step I am confused on.

Theorem. The limit of a convergent sequence in a Hausdorff space is unique.

Proof. Let $$a_n$$ be a convergent sequence in a Hausdorf space. Suppose, for a contradiction, that it converges to two different points, $$x$$ and $$y$$. Thus, it follows from converges to $$x$$ that \begin{align*} \forall \epsilon > 0, \exists N, \forall n > N, |a_n - x | < \epsilon, \end{align*} which is otherwise stated that for all $$n > N$$, elements of the sequence lie in some open ball around $$x$$ with radius $$\epsilon$$.

From convergence to $$y$$, it follows that \begin{align*} \forall \epsilon > 0, \exists N, \forall n > N, |a_n - y| < \epsilon, \end{align*} otherwise stated that for all $$n > N$$, elements of the sequence lie in an open ball around $$x$$ with radius $$\epsilon$$.

Here is where my confusion comes in. From here, I know I need to draw on the definition of Hausdorff space. These are distinct points, and so there exist open sets around them containing the points, $$x$$ and $$y$$, where these sets are disjoint. This does not imply that every open set containing these points is disjoint. So, it seems that I need to say something to the effect that the definition of convergence allows me to create an open ball (I am using this interchangeable with open set, which I hope isn't incorrect; please correct me, if so) of any radius I want around the points, and thus it clearly captures all such open sets. Thus, I can pick two separate $$N$$'s for each of these sets to form open balls of radius $$\epsilon_1$$ and $$\epsilon_2$$ around these points such that the sets are disjoint, which I know I can do via the definition of Hausdorff space. Since this would be true for an infinite number of $$n$$ past some arbitrary point $$N$$, it would not be possible to get "back inside" the opening ball around the other point. That's clearly a contradiction to the definition of convergence. Thus, if $$a_n$$ converges to $$x$$, it cannot converge to $$y$$, and if it converges to $$y$$, it cannot converge to $$x$$, so there is only a single possible limit point, which is unique.

How does this argument sound? Is there a better way to state it, or have I made an errors in logic?

• There's something you must pay attention to: when you say "Hausdorff space", a distance is not assumed to exist, and ever less a normed space structure, so you can't talk about balls of radius $\epsilon$. – Scientifica Nov 17 '18 at 23:46

Suppose $$x_n \rightarrow x$$, then if $$x \neq y$$ there exists neighbourhoods $$U,V$$ of $$x,y$$ respectively that are disjoint by Hausdorfness. Next by definition of convergence, $$U$$ must contain all but finitely many of the $$x_n$$, and so $$V$$ cannot, and so $$x_n$$ cannot converge to $$y$$, because $$V$$ is a neighbourhood of $$y$$ that does not contain all but finitely many of the $$x_n$$. I think you should use this definition of convergence of a sequence in a topological space.

I would advise you to forget about radii when studying general topology: neighborhoods are a way more general notion; they do not come in with a radius.

Anyway, remember the definition of a Hausdorff space and that's all you're gonna need: If $$x,y\in X$$ with $$x\neq y$$ then we can find $$U,V$$ open sets with $$x\in U, y\in V$$ s.t. $$U\cap V=\emptyset$$.

Okay, now suppose that $$x_n\to x,y$$ and we want to prove that $$x=y$$. Suppose that this was not true, then we can find $$U,V$$ as above; but $$x_n\to x$$ means by definition that for any open set $$A$$ with $$x\in A$$ $$(x_n)$$ is contained in $$A$$ from some index and on. Do this for $$U,V$$ and you immediately have a contradiction, since they are disjoint.

• Thank you for this answer. I think I understand the explanation, but just to be sure: beyond some $N$ (which might differ, I presume, for convergence to $x$ and $y$), every open set we construct might contain the limit point. I think where I am confused is that the definition of the Hausdorff space seems to only require the existence of a single disjoint open set. If these disjoint sets existed prior to $N$, the requirements would be satisfied. I am confident I am incorrect on this, but I cannot see why. – Matt.P Nov 18 '18 at 0:00
• Recall the definition of sequence-convergence in a topological space: we have that $x_n\to x$ if and only if for any open set $U$ with $x\in U$ there exists $n_0$ such that for all $n\geq n_0$ it is $x_n\in U$. This is independent of the Hausdorff property! The sets $U,V$ mentioned on the definition of Hausdorff spaces depend only on the points $x,y$ mentioned and nothing else! – JustDroppedIn Nov 18 '18 at 0:04
• Thank you again. One more time, if you wouldn't mind, just to be sure I'm on the right page: by the definition of Hausdorff, we take disjoint open sets $U$ and $V$ around $x$ and $y$, respectively. By convergence to $x$, we can guarantee that an infinite number of points lie within one of these, say within $U$. But, if all but a finite number of points lie within $U$, then only a finite number of points lie within $V$, which contradicts the fact that we could guarantee an infinite number of points lie within $V$. The same would be true in the opposite direction. How is that? – Matt.P Nov 18 '18 at 0:42
• @Matt.P it's good:) – JustDroppedIn Nov 18 '18 at 9:41

You are reasoning as if you're in the reals, and in the reals your argument isn't valid either as you wrote it. Just use definitions and the proof writes itself:

Suppose, for a contradiction, that for some sequence $$(x_n)$$ from $$X$$ we have $$x,y \in X$$ with $$(x_n) \to x$$ and $$(x_n) \to y$$ and $$x \neq y$$.

By the Hausdorff property, there are disjoint open sets $$U$$ and $$V$$ of $$X$$ such that $$x \in U$$ and $$y \in V$$.

As $$x_n \to x$$, and $$U$$ is an open neighbourhood of $$x$$, there is a $$N_0 \in \mathbb{N}$$, such that for all $$n \ge N_0$$ we have $$x_n \in U$$. $$(1)$$

Also, as $$x_n \to y$$, and $$V$$ is an open neighbourhood of $$y$$, there is a $$N_1 \in \mathbb{N}$$, such that for all $$n \ge N_1$$ we have $$x_n \in V$$. $$(2)$$

Now let $$m= \max(N_0, N_1)$$, then $$m \ge N_0$$ so $$x_m \in U$$ by $$(1)$$ and also $$m \ge N_1$$ so by (2) we have $$x_m \in V$$.

But then $$x_m \in U \cap V$$ contradicts the disjointness of $$U$$ and $$V$$.

This shows that all convergent sequences have a unique limit in a Hausdorff space.