# Looking for a clarification of the definition of limit of real function based on sequences

I have found the following alternative definition of limit of a real function $f \in \mathbb{R}^T$, with $T \subseteq \mathbb{R}$.

Definition: If $x$ is an extended real number which is the limit of at least one sequence in $T\setminus\{x\}$, we say that $y$ is the limit of $f$ at $x$, and write $$\lim_{t \to x} f(t) = y,$$ provided that $f(x_m) \to y$ for every sequence $(x_m)$ in $T\setminus\{x\}$ with $x_m \to x$.

I have some problems concerning it.

1. Why is it actually equivalent to the standard $\varepsilon - \delta$ definition?
2. Don't we lose something by focusing on sequences (defined on $\mathbb{N}$, while the domain here is $\mathbb{R}$)?

As always, any feedback is most welcome.

Answering $1$ and $2$ simultaneously:

It is equivalent due to the fact that $\overline{\mathbb{R}}$ is first countable: that is, for every point $x$ there exists a countable local basis. You can find more info on definitions here. Note that this is valid for an arbitrary metric space. However, the definition you state is not equivalent in general topological spaces. You can, however, solve this issue by using nets instead of sequences to characterize continuity.

• First of all, thanks a lot for the answer, because now I see the point of the definition. Still, I would like to know if/how it is possible to show the two definitions are equivalent without let first countablility enter the pictiue. Moreover, from the answer I am not completely sure about the relation between the arbitrary metric spaces and those spaces that are first countable and the definition. Could you clarify it? (i.e. do you actually mean that being a metric space implies being first countable?) – Kolmin Oct 4 '15 at 14:50
• Yes, metric space implies being first countable. Just take the balls of radius $1/n$ around a point. Actually, the standard proof of the equivalence of the claim you mention use exactly those balls. Well, one side of the equivalence is easier to see than another. For the other one, you use those balls to construct a sequence which will converge to the point, but for which the function applied to them will not converge to what is expected. You can find the proof on Rudin PMA pg. 84 – Aloizio Macedo Oct 4 '15 at 14:54
• Nice! Also, thanks a lot for the reference. I will take a look at the proof. – Kolmin Oct 4 '15 at 14:56

While Aloizio Macedo correctly states that these two definitions are equivalent because of the first countability of the reals, here is a short proof:

($\Rightarrow$) Assume that $f$ is continuous at $x\in T$ with respect to the $\varepsilon-\delta$ convention. Let $\{x_n\}$ be a sequence in $T\setminus\{x\}$ converging to $x$. Then fix $\varepsilon>0$. Then we know there exists a $\delta$ such that if $|x'-x|<\delta$, $|f(x')-f(x)|<\varepsilon$. Furthermore, we know there exists $N$ such that if $n\geq N$, $|x_n-x|<\delta$, so for $n\geq N$, $|f(x_n)-f(x)|<\varepsilon$, so $f(x_n)\rightarrow f(x)$.

($\Leftarrow$) Assume that for every sequence $x_n\to x$, $f(x_n)\to f(x)$. Suppose, to the contrary, that there exists $\varepsilon>0$ such that for all $\delta$, there exists $x'$ with $|x'-x|<\delta$, but $|f(x')-f(x)|\geq\varepsilon$. Choosing $\delta_n=1/n$, let $x_n$ satisfy $|x_n-x|<\delta$ and $|f(x_n)-f(x)|\geq\varepsilon$. Clearly, $\{x_n\}$ is a sequence converging to $x$, but $\{f(x_n)\}$ does not converge to $f(x)$. This is a contradiction to our assumption, so for every $\varepsilon>0$, there exists $\delta$ such that if $|x'-x|<\delta$, then $|f(x')-f(x)|<\varepsilon$.

Of course, the above proof does not address the issue of $x=\pm\infty$, but to do so involves only a few changes. Thus these two definitions for equality are equivalent for functions $f:T\subset\mathbb R\to\mathbb R$.

The reason we don't loose anything by considering sequences is that we are considering every sequence. It is not just enough that one sequence has the property that $x_n\to x$ implies $f(x_n)\to f(x)$, they all have to.

First countability means that at every point $x$, there exists a countable number of open sets $G_n$ such that for every open set $G$ containing $x$, $G$ also contains at least one of the sets $G_n$. This is true for all metric spaces (including $\mathbb R$), as well as the extended reals, and most topological spaces used in every day application. There are some topological spaces, though, which do not have this property, and therefore, these definitions are not equivalent. But in such spaces, we have no notion of distance between points, so $\varepsilon-\delta$ doesn't make sense either, so we have to come up with even more general characterizations of continuity.

• Thanks a lot for the answer and the short proof! – Kolmin Oct 4 '15 at 15:13