# Definition of continuity

It has been a year or so I took my course of real analysis, still could not understand these two definitions of continuity-(These two definitions are given as chapter 9 and 10 in the classroom resource material of MAA -Exploratory Examples in Real Analysis).

Definition 1:Sequence definition of continuity

$f(x_0)$ exists;

$\lim_{x \to x_o} f(x)$ exists; and

$\lim_{x \to x_o} f(x)$ =$f(x_o)$.

The sequence definition is convinient tool to prove continuity of polynomials. This definition is also useful when in proving discontinuity.

Definition 2:

Let $y=f(x)$ be a function.Let $x=x_o$ be a point of domain of $f$ .The function $f$ is said to be continuous at $x=x_o$ iff given $\epsilon \gt 0$,there exists $\delta \gt 0$ such that if $x \in (x_0-\delta,x_0+\delta)$, then $f(x)\in (f(x_o)-\epsilon,f(x_o)+\epsilon )$.

This definition is extremely useful when considering a stronger form of continuity,the Uniform Continuity.

I know that Definition 2 puts the use of $\epsilon - \delta$ ,My doubt is the use of a sequence in definition 1 ,i.e. in 1.) Given a sequence $(x_n)_{n=1}^{\infty}$ that converges to domain point $x=x_o$ of $f$ ,we are interested in determining the relationship between the continuity of $f$ at $x=x_o$ and the behaviour of corresponding sequence of outputs $(f(x_n))_{n=1}^{\infty}$ which forms the basis of the definition. How is it equivalent to 2.) Is there some proof for this..

Any help would be appreciated.

• Definition 2 just replace the limit in definition 1 by definition of limit. – user175968 Sep 17 '14 at 11:29
• I think the third 'definition' is true too. What it means for a function to be continuous at $x_0$ is (intuitively speaking) by moving $x$ closer to $x_0$ you should be able to get $f(x)$ arbitrary close to $f(x_0)$. (There should be no 'gap'.) This helped me understand those concepts. – flawr Sep 17 '14 at 11:31
• As one of the answers points out, merely having some sequence of $x_n$ for which $f(x_n)$ converges is not sufficient. But the statement $\lim_{x \to x_0} f(x) = f(x_0)$ is much stronger than merely a statement about a single countable sequence. – David K Sep 18 '14 at 0:22
• @DavidK could you please elaborate on how $lim_{x \to x_o} f(x)=f(x_o)$ is stronger .. – spectraa Sep 18 '14 at 1:26
• @WantTobeAbstract See math.stackexchange.com/questions/318921/limit-of-a-function for an example of how to define limit of a function; it looks a lot like your definition 2, in fact. On the other hand you already have received an answer explaining how a sequence of function values can converge on $f(x_0)$ although the function is not continuous there (in fact, $\lim_{x \to x_0} f(x)$ does not exist). – David K Sep 18 '14 at 3:20

Claim: A function is continuous if and only if it sends all converge sequence to converge sequence, limit to limit.(Here we restrict our discussion on function on $R$)

=>
$f$ is a continuous function
$a_1,a_2,...$ is a converge sequence that converge to a.
For the sequence $f(a_1),f(a_2),...$ and a given $\epsilon$, because f continuous, there exist $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$. For that $\delta$, because $a_n$ converges to $a$, there exist $N$ such that $n\geq N$ implies $|a_n-a|<\delta$, which imples $|f(a_n)-f(a)|<\epsilon$. Therefore $f(a_n)$ converges to $f(a)$.
<=
If $f$ is not continuous at given $a$, then there exist an $\epsilon>0$, for all $\delta>0$ we can find $x$ such that $|x-a|<\delta$ but $|f(x)-f(a)|\geq \epsilon$. Take $\delta=1,1/2,1/3,1/4,1/5,...$ and find the corresponding $x$ it forms a converge sequence but it's image does not converge to $f(a)$ (distance always greater or equal to fixed $\epsilon$).

I believe in order to write a proof, one needs to be able to visualize what they are trying to prove mentally.

So here is an illustration I made for definition 2

Let $y=f(x)$ be a function.Let $x=x_o$ be a point of domain of f .The function f is said to be continuous at $x=x_o$ iff given $\epsilon \gt 0$,there exists $\delta \gt 0$ such that if $x\in (x_o−\delta,x_o+\delta )$, then $f(x)\in (f(x_o)−\epsilon ,f(x_o)+\epsilon )$. And here is an illustration I made for definition 1

$f(x_0)$ exists;

$\lim_{x \to x_o} f(x)$ exists; and

$\lim_{x \to x_o} f(x)$ =$f(x_o)$. • @oa-archy Although my question was regarding the continuity definition and the sequences whose answer frankooo gave.But this was a nice intutive way & it helped.Thanks. – spectraa Sep 17 '14 at 11:55
• @WantTobeAbstract Glad to hear that. I was making the second illustration back then. – 吖奇说 ARCHY SHUō Sep 17 '14 at 11:58

Hint: The definition of $$\lim_{x\to x_0} = a_0$$ is the following:

For every $\epsilon>0$, there exists a $\delta > 0$ such that for every $x$ for which $|x-x_0|<\delta$ is true, the statement $$|f(x)-a| < \epsilon$$ is also true.

• NB: You only need one > to start a quotation environment ;) – AlexR Sep 17 '14 at 11:34

Definition 2 supposes that $f(x_0)$ exists, since $x_0\in f$ domain... But after this, you have one version that details the concept of limit, not the other one...

As for the series and the continuity, beware...

Let's have $x_n=\frac{1}{n}$ and $f$ such as $f(x)=x²$ for $x=1/q$, $q$ integer, and $f(x)=x$ for all other $x$/

Then you have $lim_{n \to \infty}x_n=x_0$ and $\lim_{n \to \infty} f(x_n)=f(x_0)=0$, but $f$ is not continuous...

• you mean to say that definition 1.) is not sufficient to prove continuity but definition 2.) does. Please explain .I already upvoted for you.. – spectraa Sep 17 '14 at 13:09
• also please explain the second line in answer. – spectraa Sep 17 '14 at 13:12
• @WantTobeAbstract, I am saying that definition 1 and 2 are equivalent. 2 is more explicit than 1, for it details the definition of the limit. – Martigan Sep 17 '14 at 13:42
• @WantTobeAbstract. For the second line of the answer, I am saying that to prove that a function have a limit on a serie does not mean that the function is continuous for all x... – Martigan Sep 17 '14 at 13:45
• To be more precise, the only thing you can prove is that $f$ continuous means that $f(x_n)$ converges when $n \to \infty$, but the reverse is not true. – Martigan Sep 17 '14 at 13:47

The two definitions are equivalent. If you comemorate the definition of a limit: $$\lim_{n\to\infty} a_n = a$$ is defined by:

For all $\epsilon > 0$ there exists $N\in\mathbb N$ such that $|a_m - a| <\epsilon \ \forall m > N$.

You should easily see their equivalence now.