# Is proving sequential continuity more difficult than proving continuity?

For a function $f: \mathbb{R}^n \to \mathbb{R}^m$, I know that continuity and sequential continuity are equivalent.

Sequential continuity of $f$ at a point $x \in \mathbb{R}^n$ means for any sequence in the domain that converges to $x$, the function values on the sequence also converge to $f(x)$. I feel this definition difficult to apply even when the domain is $\mathbb{R}$ or its subset, because how can one possibly consider all sequences that converge to $x$? There are many cases that a sequence can converge to $x$.

On the other hand, continuity of $f$ at a point $x \in \mathbb{R}^n$ means for any neighbourhood of (or just open ball centered at) $f(x)$, there exists a neighbourhood of (or just open ball centered at) $x$, whose image under $f$ is contained in the one of $f(x)$. I feel this definition is easier to apply, but I don't know how to explain why.

So is my understanding of proving sequential continuity more difficult than proving continuity right?

If needed, here is an example: prove $f([x,y])=x-\sqrt{y}$ is continuous at $(1,1)$.

Thanks and regards!

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Note that any sequence that converges to $x$ will eventually lie in any given open neighborhood of $x$... And if your objection lies in that there are "a lot" of sequences that converge to $x$, why does it not bother you at all that there are a lot of open balls centered at $f(x)$? You need to consider all of them! –  Arturo Magidin Feb 8 '12 at 6:04
@ArturoMagidin: Thanks! Do you feel one way may be more complicated than the other for some other reason? For example, when domain is $\mathbb{R}$, do you have preference of which way? –  Tim Feb 8 '12 at 6:09
I don't think either one is more complicated per se; they often (usually) depend on the particular function. –  Arturo Magidin Feb 8 '12 at 6:17
In my experience, showing that a function is NOT continuous can be easier with sequences, here you only need to find one sequence at some discontinuity point. But when showing that a function is continuous, I prefer the $\varepsilon - \delta$ method (or nghoods incase more general setting than metric spaces is considered). –  Thomas E. Feb 8 '12 at 6:40
Sequential continuity feels (to me) more concrete than continuity. Indeed it is almost a theorem that it is more concrete, since in general we need the Axiom of Choice to conclude continuity from sequential continuity. –  André Nicolas Feb 8 '12 at 7:10