# 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). – T. Eskin 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

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

It may be this depends on how you think about sequences. If you are happy working with sequences in the abstract sense, the two definitions will feel very similar. If you think in terms of specific sequences, you will find it hard to write down a proof that works for each one.

There is an argument to me made for the opposite conclusion:

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

In practice:

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

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