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

Question

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.
Thank you for your time.

Answer 1

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.

Answer 2

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.