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 ?

  • 1
    $\begingroup$ You want punctured neighborhoods (neighborhood of $x$ that exclude $x$ itself). $\endgroup$ Nov 4, 2021 at 13:38

1 Answer 1


You're right in saying that I should be more careful there and take the filter base consisting of all punctured neighborhood of the given point (and the corresponding filter).

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 $x\to\infty$).

The definition of limit which takes all neighborhoods (not only the punctured ones) is sometimes used, but it is definitely rarer. See also: What's the purpose of the two different definitions used for limit?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .