# "usual limit" from filter

I'm reading this in which the author says that $$\lim_{x\to a}f(x)=y$$ if for every neiborhood $$U$$ of $$y$$, $$f^{-1}(U)$$ belongs to the filter of neigbohoods of $$a$$.

What about the following function ? $$f(x)=1$$ if $$x=0$$ and $$f(x)=0$$ otherwise.

With the usual notion of limit, we have $$\lim_{x\to 0}f(x)=0$$ regardless to the fact that $$f(0)$$ itself has a "strange" value.

With the limit defined from the filter of neigborhoods, the function $$f$$ has no limit at $$x=0$$.

EDIT(add justification): if $$U=(-1/2, 1/2)$$ we have $$f^{-1}(U)=R\setminus\{0\}$$, which is not a neigborhood of $$0$$.

Am I missing something or the "filter" limit does not brings back the usual notion of limit ?

• You want punctured neighborhoods (neighborhood of $x$ that exclude $x$ itself). Nov 4, 2021 at 13:38

I will add that, for example, Dixmier's General Topology uses the same notion of limit along a filter base (Definition 2.2.1) and the author takes this as the basic notion of the limit - while the limit corresponding to punctured neighborhoods is taken as one of the variants listed in Example 2.2.4 (together with some other kinds of limit, such as $$x\to x_0^+$$ or $$x\to\infty$$).