Uniqueness of the limit of a function for Hausdorff spaces that are not metric spaces Is there a proof to show that a mapping from an arbitrary topological space to an arbitrary Hausdorff space (excluding metric spaces) has a unique limit point? Therefore, the only properties to be used are:
1) inverse continuous mappings of open sets are open sets
2) any two points in a Hausdorff space can have disjoint open neighborhoods
3) any point in a Hausdorff space is a close set
Hints will be appreciated. 
 A: In a general topological space, the appropriate notion of convergence rely on the notion of filter base.
Let $X$ a set. A filter $F$ on $X$ is a set of non-empty subsets of $X$ such that $A,B\in F$ implies $A\cap B\in F$, and $A\in F$ and $A'\subseteq A$ implies $A'\in F$. A filter base on $X$ is a set $F$ of non empty subsets of $X$ such that for each $A,B\in F$, there exists $C\in F$ such that $C\subseteq A\cap B$. (In particular $A\cap B\not=\varnothing$.) Note that a filter is a base filter. (The reciprocal is false in general.)
If $X$ is a topological space and if $x_0\in X$, the set $V(x_0)$ of neighbourhoods of $x_0$ in $X$ is a filter. Any fundamental system of neighbourhoods of $x_0$ in $X$ is a filter base on $X$.
If $X$ is a topological space, if $Y\subseteq X$ and if $x_0 \in \overline{Y}$, then the set $F_{x_0}(Y)$ of subsets of $Y$ of the form $Y\cap V$ where $V$ is a neighbourhood of $x_0$ in $X$ is a filter on $Y$. ($Y\cap V\not=\varnothing$ especially because $x_0 \in \overline{Y}$.) If $Y = X$ you recover the previous definition.
The more general case of limit of a function is the following : let $X$ be a set and $F$ a filter base on $X$, and $f : X\rightarrow Y$ a map where $Y$ is a topological space, and $l\in Y$. One says that $f$ tends to $l$ along $F$ if for each neighbourhood $V$ of $l$ in $Y$, there exists a $B\in F$ such that $f(B)\subseteq V$. Then, we have the unicity theorem :
Theorem. Let $X$ be a set and $F$ a filter base on $X$, and $f : X\rightarrow Y$ a map where $Y$ is an Hausdorff topological space, such that $f$ has a limit along $F$. Then this limit is unique
Proof Let $l,l'\in Y$ be limits of $f$ along $F$. As $Y$ is Hausdorff, you can find two neighbourhoods $V$ resp. $V'$ of $l$ resp. $l'$ in $Y$ such that $V\cap V'= \varnothing$. Then there exists $B,B'\in F$ such that $f(B)\subseteq V$ and $f(B')\subseteq V'$. As $F$ is a filter base, there is a $B''\in F$ included in $B\cap B'$, then $f(B'')\subseteq V\cap V'$. As $B''\not=\varnothing$ (an filter is constitued of non-empty sets...) necessarily $V\cap V'\not= \varnothing$, a contradiction.
Now, take $X = Y = \mathbf{R}$. Thanks to the previous notions, we can handle the following notions in the same way, thanks to (different filter bases) :


*

*$\lim_{x\rightarrow x_0} f(x)$ (this nototion is used in the definition of continuity at $x_0$)

*$\lim_{x\rightarrow x_0, x\not = x_0} f(x)$ (this notion is used in the definition of limit of a fucntion at some point)
$\lim_{x\rightarrow x_0} f(x) = l$ means that $f$ as limit $l$ along the filter (a filter is a filter base, remember) $V(x_0)$, whereas $\lim_{x\rightarrow x_0, x\not = x_0} f(x) = l$ means that $f$ has limit $l$ along the filter base $F_{x_0}(Y)$ with $Y = \mathbf{R}\backslash\{x_0\}$. (These are in fact definitions)
Last remark : $x_0 = 0$ and if $f$ is the map equal to zero on $Y$ and $1000$ at $x_0$, one should take care of the fact that $\lim_{x\rightarrow x_0} f(x)$ does not exist because $f$ si not continuous at $x_0$, whereas $\lim_{x\rightarrow x_0, x\not = x_0} f(x)$ does and is equal to $0$.
