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.

  • $\begingroup$ What is the definition of "limit point of a mapping"? $\endgroup$ – Krish Jan 18 '15 at 18:43
  • $\begingroup$ I think the OP means an (existing) limit of $f$ at some point $x$. An wants to know why the "an" could be replaced by "the" in case of Haussdorfness of the target of $f$. $\endgroup$ – Olórin Jan 18 '15 at 18:47

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 $0$.

  • $\begingroup$ Btw, remark that this proof is valid for metric spaces as well. Haussorfness is NOT a metric property, but a topological one, as it simply means that the diagonal in the fiber product $X\times_Y X$ is closed (topological property) in the fiber product $X\times_Y X$. (Here I noted $f : X\rightarrow Y$ the map.) The fiber product $X\times_Y X$ is just the subset of $X\times X$ consisting in couples $(x,x')$ such that $f(x) = f(x')$. $\endgroup$ – Olórin Jan 18 '15 at 18:49
  • $\begingroup$ How does the fact that "As $l,l'$ are limits of $f$ at a point $x$" show that "you can find a neighbourhood $U$ (resp. $U'$) of $x$ such that $f(U)\subseteq V$ (resp. $f(U')\subseteq V'$)."? $\endgroup$ – Student Jan 18 '15 at 20:20
  • $\begingroup$ By definition, $f : X \rightarrow Y$ converges to $l\in Y$ at $x\in X$ if and only if for each neighbourhood $V$ of $l$ in $Y$ there exists a neighboorhoud $U$ of $x$ in $X$ such that $f(U)\subseteq V$. Think of same definition it the case where $X = Y = \mathbf{R}$ with the usual topology... $\endgroup$ – Olórin Jan 18 '15 at 20:21
  • $\begingroup$ Here is a counter example. Consider the mapping $f : [-1,1] \rightarrow R$and $f(x) = x^2$ for all $x \not= 0$ and $f(0) = 1000$. Then as $x \rightarrow 0$ $f(x) \rightarrow 1000$ and so 1000 is a limit point. But no open neighbourhood of $0$ can be mapped to a subset of the point 1000. $\endgroup$ – Student Jan 18 '15 at 20:23
  • $\begingroup$ What you have stated is the condition that $f$ is continuous. In this case it need not be continuous. $\endgroup$ – Student Jan 18 '15 at 20:25

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