For all my homework in real analysis, when I've been asked to show that a function is continuous, I just found a single $x_n \in D$ and showed that when $x_n \rightarrow x_0$, $f(x_n) \rightarrow f(x_0)$. Apparently, the sequence definition (as opposed to the epsilon delta definition) is (basically) only used to prove a function is not continuous, and I can't prove a function is continuous because then I'd have to show this is true for all possible sequences? Am I doing the math wrongly? Should I always use the epsilon delta definition when trying to prove that a function is continuous?

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    $\begingroup$ That would be incorrect. $\endgroup$ – Math1000 May 27 at 0:53
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    $\begingroup$ oh god. this is not good. $\endgroup$ – MinYoung Kim May 27 at 0:54
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    $\begingroup$ The sequential defintion says that for all sequences $(x_n)$ such that $x_n\to x$, we have $f(x_n)\to f(x)$. That means that to prove a function is continuous, it is not enough to work with a single sequence, you need to prove that any sequence that converges to $x$ will “work.” By contrast, the negation of this is “there exists a sequence $(x_n)$ such that $x_n\to x$, but $f(x_n)$ does not converge to $f(x)$”. So to prove, using sequences, that a function is not continuous, you only need to exhibit a single sequence where things go haywire. $\endgroup$ – Arturo Magidin May 27 at 0:55
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    $\begingroup$ Basically, if I said “Everyone here is named Todd”, then to prove me wrong it would be enough for someone to stand up and say “My name is Charlie.” But to prove me right, it is not enough for someone to stand up and say “That’s right: my name is Todd.” $\endgroup$ – Arturo Magidin May 27 at 0:55
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    $\begingroup$ @MinYoungKim: The key word in the definition you quote is whenever, which generally means “for all (things) such that…”, unless some explicit restriction on the things has been previously given. So written out a bit pedantically, “whenever $x_n \to x_0$” means “for all sequences $x_n$ and points $x_0$ such that $x_n \to x_0$”. $\endgroup$ – Peter LeFanu Lumsdaine May 27 at 10:29

This is indeed incorrect. Take for example the function $$f(x) = \begin{cases} 1, \text{ if } x\in \mathbb{Q}\\ 0, \text{ otherwise} \end{cases}$$

This is obviously not a continuous function. However, if you look at its behavior along a sequence of rational points, it would appear to be constant (hence continuous).

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    $\begingroup$ I am baffled by the popularity of my answer. $\endgroup$ – tia May 30 at 11:41
  • $\begingroup$ I just followed the crowd $\endgroup$ – MinYoung Kim May 31 at 4:20

Yes, unfortunately proving continuity requires showing for every sequence $x_n$ if $x_n\to x$ then $f(x_n) \to f(x)$

One specific sequence does not prove continuity.


What everyone said is true. But I'd like to say a few more things. Like what the other commenters wrote, if you want to prove $f:D \rightarrow \mathbb{R}$ is continuous, you need to say "Let $x_0 \in D$ and let $(x_n)$ be a sequence in $D$ that converges to some $x_0$." $(x_n)$ in this proof is an abstract concept: it's simply an arbitrary sequence in $D$. It's not anything special or specific, it's just some regular sequence that happens to converge to $x_0$.

From there, you have to use the mathematics of sequences, convergence, properties given to you on the problem to logically walk from the statement "$\lim x_n = x_0$" to "$\lim f(x_n)=f(x_0)$." Remark you can't use the properties of any specific sequence, such as $(x_0+\frac{1}{n})_n$.

You also asked if you should always use $\epsilon-\delta$ definitions. So remember I implied you have to play around with the mathematics of sequences, convergence, etc. for arbitrary sequences. This same goes for $\epsilon-\delta$ definitions. Trying to prove a function is continuous by using an actual specific sequence is like trying to prove a function is continuous by setting $\epsilon=1$ and showing there exists some $\delta>0$ such that $|f(x)-f(x_0)|<1$ whenever $x \in D$ and $|x-x_0|<\delta$. Like...congrats. You did it for $\epsilon=1$, but you didn't do it for $\epsilon=2$. You didn't do it for $\epsilon>0$. It's just for $\epsilon-\delta$ definitions, the idea of using arbitrary $\epsilon$'s and $\delta$'s is really obvious compared to the sequence definition. However the idea is still the same: you have to be abstract and arbitrary. I hope this helped.


Is not enought find a particular sequence $x_{n}$ such that $x_{n} \to x_{0} \implies f(x_{n}) \to f(x_{0})$ Well, $f:[0,2] \to \mathbb{R}$ such that $f(x) = 0$ if $ x\in [0,1] $ and $f(x) = 1 $ if $x \in (1,2]$ holds if we take $x_{n} = 1-\frac{1}{n}$ then $x_{n} \to 1$ and $ f(1-\frac{1}{n}) \to f(1)$ but $f$ is not continuos in $x = 1$


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