In Rudin's functional analysis text, he defines convergence in Hausdorff spaces in the following way:

$x_n\rightarrow x$ if every neighborhood of $x$ contains all but finitely many $x_n$.

Is this the definition for convergence in topological spaces generally? If so, any idea why he would specify Hausdorff spaces? I mean I supposed the definition is only clearly meaningful in Hausdorff spaces since this kind of bunching in neighborhoods of $x$ is only 'special' in a space in which we are capable of separating points into disjoint neighborhoods. Is that the only thing going on here? Is there a better way to understand convergence in non-Hausdorff spaces?

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    $\begingroup$ I think Rudin makes the distinction because in Hausdorff spaces, the limit will be guaranteed to be unique. In non-Hausdorff spaces, it may be that every sequence with infinitely many distinct points "converges" to every point in the space (think of $\Bbb R$ with the cofinite topology). $\endgroup$ – Clayton Mar 19 '19 at 13:33
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    $\begingroup$ The definition of $x_n\to x$ works in any topological space, Hausdorff or not. But people often use the alternative notation $\lim_{n\to\infty}x_n=x$, and this notation should be restricted to Hausdorff spaces. In non-Hausdorff spaces, you could have $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}x_n=y$ and yet $x\neq y$, contrary to how the symbol "$=$" should behave. $\endgroup$ – Andreas Blass Mar 19 '19 at 19:17

This is the definition of convergence of sequences in general topological spaces, however sequences aren't always sufficient when talking about convergence. The usual definition for convergence of a sequence in general topological spaces is the following:

Definition: A sequence $(x_{n})$ in a topological space $X$ is said to converge to a point $x\in X$, if for every open neighbourhood $U\subseteq X$ of $x$ there is an $N\in\mathbb{N}$ such that for every $m\geq N$ one has that $x_{m}\in U$.

This is exactly the definition you gave in your question. However, if your topological space is not Hausdorff, then a sequence may converge to multiple points. For example let $X$ be the underlying set of $\mathbb{R}$ and equip it with the indiscrete topology. That is, the only open sets are $\mathbb{R}$ and $\emptyset$. Then given any sequence $(X_{n})$ in $X$ we have that $(x_{n})$ converges to every point of $X$.

If $X$ is assumed to be Hausdorff then a sequence can converge to at most one point.

Prop: Let $X$ be a Hausdorff space and $(x_{n})$ a sequence that converges to the points $x$ and $y$. Then $x=y$.

If $x\neq y$ then there are disjoint open sets $U,V\subseteq X$ that contain $x$ and $y$ respectively, say $x\in U$. Because $(x_{n})$ converges to $x$ we have that $U$ contains all but finitely many of the $x_{n}$. This implies that there are infinitely many of the $x_{n}$ that are not in $V$, an open neighbourhood of $y$. This contradicts $(x_{n})$ converging to $y$. Therefore $x$ and $y$ must be the same.

Interestingly enough, convergence of sequences is, in general, insufficient when talking about limits. By this I mean that if $X$ is a topological space and $x\in X$ is a limit point of a set $A\subseteq X$ then you can not say that there is a sequence $(x_{n})$ in $A$ that converges in $X$ to $x$. An easy example is the the set $X=\omega_{1}+1$, the set of all countable ordinials together with the first uncountable ordinal, equipped with the order topology. The element $\omega_{1}$ is a limit point of the set of countable orders in this topology, but there is no sequence of countable ordinals in $X$ that converges to $\omega_{1}$.

Even worse (better) there are spaces where limits of ordinal indexed sequences are not enough. I believe that the token example is the Tychonoff plank in which there is a an element that is in the closure of a set, but no ordinal indexed sequence in that set converges to that point.

The most general notion of convergence within a space is that of the convergence of nets. In a topological space $X$, a point $x\in X$ is a limit point of a set $A\subseteq X$ if and only if there is a net in $A$ that converges to $x$. Instead of adding more definitions to this already long post, I will refer you to wikipedia.


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