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?
Thanks in advance.
Best Answer
Suppose $x_n \rightarrow x$, then if $x \neq y$ there exists neighbourhoods $U,V$ of $x,y$ respectively that are disjoint by Hausdorfness. By the 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, given in Munkres Topology.