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Discontinuity Criterion Let $A \in \mathbb{R}$, let $f : A \to \mathbb{R}$, and let $ c \in A$. Then f is discontinuous at c if and only if there exists a sequence $(x_n)$ in A such that $(x_n)$ converges to c, but the sequence $(f(x_n))$ does not converge to $f(c)$

I'm having trouble understanding this definition. With the divergence criteria, we pick out any sequence that converges to c such that it isn't equal to c. However, over here, it says that the sequence is in A. What does that mean? How does that change the way I pick out sequences. Moreover, I don't think I get the motivation behind the sequential criteria. Can anyone give me a geometric interpretation(or any other for that matter) of this so I can see how functions and their convergence and continuity is related to that of sequences?

Thank you.

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  • $\begingroup$ Do you know the corresponding statement about continuity? That $(x_n)$ is a sequence in $A$ just means that $x_n \in A$ for any $n$. $\endgroup$ – Gibbs Apr 14 '18 at 20:30
  • $\begingroup$ Do you mean the definition of continuity? If yes, then I do know it. $\endgroup$ – A.Asad Apr 14 '18 at 20:38
  • $\begingroup$ That is not really the definition of continuity of a function. It is an equivalent statement (see Definition in terms of limits of sequences, there is even a picture explaining the result for the exponential function). If you know that, the statement in your question is just the opposite. $\endgroup$ – Gibbs Apr 14 '18 at 20:43
  • $\begingroup$ If $x_n$ wasn’t in $A$. $f(x_n)$ wouldn’t exist. Moreover, the sequence must converge to an element $c$ of $A$, otherwise $f(cj$ wouldn’t exist. (A remarkable consequence is that every function defined on a subset of the integers is continuous.) $\endgroup$ – Michael Hoppe Apr 14 '18 at 21:26
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I'll try to give a good example (as intuitive as I can) to help you understand this definition.


Define the function $f:[0,2]\rightarrow\mathbb{R}$ $$ f(x)= \begin{cases} 1,& 0\leq x< 1 \\ 1000,&x=1\\ 3,& 1< x\leq 2\\ \end{cases} $$ And note that $f$ is not continuous at $x=1$. Now, define a sequence $(x_{n})_{n=1}^{\infty}\subseteq[0,2] $ by $$x_{n}=1-\dfrac{1}{n}$$ for all positive integers $n\in\mathbb{N}$. Then you can easily verify that $\lim\limits_{n\rightarrow \infty}x_{n}=1$. But now, observe that when you take the limit of $f$ evaluated at each $x_{n}$ you can see that

$$\lim\limits_{n\rightarrow\infty}\bigl\{f(x_{n})\bigr\}=\lim\limits_{n\rightarrow\infty}\Bigl\{f\bigl(1-\frac{1}{n}\bigr)\Bigr\}$$ For me personally, I think that it helps to think of the actual terms in this sequence. The first few are $$\Bigl\{f(0),f\Bigl(\frac{1}{2}\Bigr),f\Bigl(\frac{2}{3}\Bigr),f\Bigl(\frac{3}{4}\Bigr),\dots\Bigr\}= \Bigl\{1,1,1,1,\dots\Bigr\}$$ So $\lim\limits_{n\rightarrow\infty}x_{n}=1$, but clearly $\lim\limits_{n\rightarrow\infty}\bigl\{f(x_{n})\bigr\}=1\neq 1000$.


Now, why do we care? Why not just negate the $\epsilon-\delta$ definition of continuity to characterize discontinuity? The answer to this is that it is often much easier to show a function is not continuous at a point using this Sequential Criterion for discontinuity than it is to use the negation of the $\epsilon-\delta$ definition. If you want to convince yourself, try this as an exercise. Consider the function $$g(x)= \begin{cases} \sin\bigl(1/x\bigr),&x\neq 0\\ 0,& x=0 \end{cases} $$ on any interval containing $0$ and try to produce a single $\epsilon > 0$ such that for every $\delta>0$, there exists $x\in(-\delta,\delta)$ with $0<|x-0|<\delta$ and $|g(x)-g(0)|$. And after a minute or two, just make a sequence $(x_{n})_{n=1}^{\infty}$ which converges to $0$, such that $g(x_{n})$ doesn't converge to $g(0)$, and you are done.


The Sequential Criterion for Discontinuity also constitutes a beautifully short proof that Dirichlet's Function defined by $$g(x)= \begin{cases} 0, &x\in\mathbb{R}\setminus\mathbb{Q}\\ 1, &x\in\mathbb{Q}\\ \end{cases} $$ Is discontinuous at every rational number in $\mathbb{Q}$ (Just construct a sequence of rational numbers which converge to an irrational number).

NOTE. Discontinuity of $g$ can also be proved using the Archimedean Property.


Hope this helps, let me know if you have any questions

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  • $\begingroup$ Thank you! This helps a great deal! Beautifully explained. Can you please elaborate/hint on how to prove the discontinuity of g using the Archimedian property? $\endgroup$ – A.Asad Apr 15 '18 at 6:23
  • $\begingroup$ I should have elaborated, but i meant the version of the Archimedean property that is essentially equivalent to the density of the rationals in the real numbers. Its the most common method of proof. I'm sure it's already been asked $\endgroup$ – JB071098 Apr 16 '18 at 1:11

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