Sequential definition of continuity: What does "all sequences mean
I've been exposed to both the classic and sequential definition of continuity. The sequential definition is the following:
A function $f: A \to \mathbb R$ is continuous at a point $a \in A$ if for each sequence $(x_n)$ in $A$ such that $x_n \to a$, we have $f(x_n) \to f(a)$.
I've watched some videos on this and also read other questions, but I'm not fully onboard with this definition yet.
What's clouding my understanding is "... if for each sequence $(x_n)$ in $A$ such that $x_n \to a$ ...". In particular:
- What does "each sequence" (in the interval $A$) actually mean? My understanding is that we want to ensure there are no jumps/gaps in the graph of $f$, thus we'd need to check the output of $f$ for every real number on the $x$-axis to give a point on the $y$-axis - which is of course not possible - but how does "every sequence in $A$" cover all real numbers in the interval on the $x$-axis?
- When I look at a graph of for example the topologist's sine function, and we can focus on the interval $(-1, 1)$ and we let $a=0$ - then I intuitively think about two sequences on the $x$-axis: the sequence of real numbers approaching $0$ from the left and one approaching $0$ from the right (just "following" the $x$-axis in my mind towards $0$.) And, intuitively, I want to verify that both tend or not tend to $f(a)$ according to the definition. But - $f(0)$ is not even defined, since that'd be a division by $0$. How does this then fit with the definition?
3 Answers
$\begingroup$Great questions!
(1) Each sequence means that, no matter what sequence $\left(x_n\right)_{n \geq 1}$ you pick, if $x_n \to a$, then $f(x_n) \to f(a)$. I would personally recommend reading a proof that the limit definition is equivalent to the sequential definition (Wikipedia has a proof, as will any analysis textbook). Working through that proof in detail may help you understand why this makes sense. I can still try to provide an intuition.
The very vague intuition for continuity is that, as $x$ approaches to $a$, you want $f(x)$ to approach $f(a)$. For the sequential definition, I think it'll actually be more helpful to picture not a sequence of inputs and a sequence of outputs seperately, but rather the sequence $\left(x_n, f(x_n)\right)_{n \geq 1}$. That is, picture the points in $\mathbb{R}^2$ instead of thinking about seperate sequences along the $x-$axis and $y-$axis. What the sequential definition is saying is that, if you plot such a sequence of points, if the $x-$coordinates approach $a$, then the $y-$coordinates must also approach $f(a)$. I almost think of this as sampling points on the function - if you pick a specific sequence which converges to $a$, graph your function, and delete all the points except those where the $x-$coordinate is a term in your sequence. For an "obviously" continuous function, clearly no matter which points you delete, if the $x-$coordinates are approaching $a$, the $y-$coordinate must be approaching $f(a)$. But, for a discontinuous function, you can get different things depending on what you delete. For a function like $f(x) = \sin\left(\frac{1}{x}\right)$, if you delete everything but the peaks you get a different limit than if you delete everything but points where the graph intersects the x-axis, so it's discontinuous. For a function with a jump, if you come from the left it looks different than if you come from the right. For a removable discontinuity, the limit of the points won't be the value of the function, just as a consequence of how limits work. So, for discontinuous functions, you can find sequences that disagree with each other or with the value of the function; therefore, if all sequences agree with the value of the function and each other, the function must be continuous. I know this was vague but I hope that it's intuitive - if not, pick some functions, and pick specific sequences. Working through numeric examples can be very helpful.
(2) You've come across a pretty subtle point that most calculus/analysis students run into at some point! The thing here is that a function is actually only continuous at a point that is in its domain. The definition of continuity at some point $a$ requires that $f(a)$ is defined - this isn't just true of the sequential definition. Now, the topologists sine curve is defined at $0$; it is defined by the function
$$f(x) = \begin{cases}\sin\left(\frac{1}{x}\right) & & x \neq 0 \\ 0 & & x = 0 \end{cases}$$
so in this case there's no issue. But, if the topologists sine curve isn't defined at $0$, it isn't discontinuous at $0$, the same way that it isn't continuous by the technical definition, since $f(0)$ doesn't exist. It is however very common to abuse language/notation to say that a function is not continuous at a point $a$ outside the domain if no value of $f(a)$ makes the function continuous. In this way, the function $f(x) = \sin\left(\frac{1}{x}\right)$ is discontinuous at $0$ - there is no value of $f(0)$ that makes it continuous, and this can be seen intuitively because of the rapid oscillation.
$\endgroup$ $\begingroup$The point is that when you say you don't want a "jump" in your function, you are actually meaning that you can get as close to $f (a)$ as you want (provided you remain close enough to $a$). This is the basic crux of the definition(s) of continuity. One of the standard ways to define continuity is the $\epsilon$-$\delta$ definition, which precisely tells us the first statement I made. As for the sequential definition, let us try to understand what we want, first!
We want that whenever we are close enough to the point $a$ in the domain (in your case $A$), the image is also close enough to $f \left( a \right)$. Now, the question is, how can we get close enough to $a$? One way is to choose a small (enough) interval, say $\left( a - 1, a + 1 \right)$ (here, I consider the length of $2$ to be "small") and then choose a point in this interval. Look at the image of this point. Is the image "close enough" to $f \left( a \right)$? Even if it is, we cannot conclude from here that the function is continuous (as you said, we have to check for "all real numbers close to $a$"). So, make a smaller interval and choose another element. Again look at its image. Ask the same question. Keep on doing this by decreasing the size of the intervals. What do we get? A sequence which converges to $a$. Along with this, we also get a sequence in the codomain. Now, if we want that the images be close enough to $f \left( a \right)$ (for all real numbers close to $a$), it is necessary that the image sequence (corresponding to the sequence we constructed) has to converge to $f \left( a \right)$.
Can we say now that the function is continuous? No! Because there are still a lot of real numbers around $a$ which we have not checked. How to do it? Construct another sequence and do the same thing all over again. And keep on constructing sequences until you have exhausted all possible ways (and hence all real numbers close enough to $a$) to reach $a$. This is way the phrase "for each sequence" is required in the definition.
To talk about continuity at a point, you need the function to be defined at that point. Since in your case, $f \left( 0 \right)$ is not defined, you cannot talk about the continuity of $f$ at $0$. However, this is an interesting exercise. Define $f \left( 0 \right)$ to be some real number. Now, will the function be continuous? Notice that the oscillations are increasing as we approach zero (on the $X$-axis). Therefore, we can always find a sequence whose image sequence is far away from any real number.
- The phrase "for each sequence in $A$" means that you can take any sequence $x_1,x_2,x_3, \dots$ so that all the $x_i$'s are in $A$, and if this sequence converges to $a$, then the sequence $f(x_1),f(x_2),f(x_3),\dots$ should converge to $f(a)$.
For example, the function $f(x) = 2x^2+3$ is continuous at $x=0$. For example, let $A$ be the interval $[-1,1]$. One possible sequence that converges to $0$ is the sequence $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots, \frac{1}{n}, \dots$. If you evaluate the function at these values, you get the sequence $3+\frac{1}{2}, 3 + \frac{2}{9}, 3 + \frac{1}{8}, \dots, 3+\frac{2}{n^2}, \dots$. This sequence certainly converges to $3$ (which is $f(0)$).
But of course that isn't the only sequence which converges to $0$. You could also have the sequence $\frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \dots, \frac{1}{n^2}$, and this also converges to $0$. Evaluating the function $f$ at these values gives the sequence $3 + \frac{2}{n^4}$, which again converges to $3$.
And then of course there are lots of other sequences which converge to $0$ (I only gave two examples). So the sequential definition of continuity in this example says that for every possible sequence which converges to zero, the corresponding sequence $f(x_i)$ (where you plug the values of the sequence in to the function) converges to $3$.
- How does the topologist's sine curve fit with this definition?
This function is not continuous at $a=0$. The reason is that we can find two different sequences which both converge to $0$, but where the limits of sequences $f(x_i)$ are different. For example, take for the first sequence$$ \frac{1}{\pi}, \frac{1}{2\pi}, \frac{1}{3\pi}, \frac{1}{4\pi}, \dots, \frac{1}{n\pi}, \dots $$This sequence certainly converges to $0$. Since the function is $f(x) = \sin(1/x)$, the corresponding $f(x_i)$ sequence is $\sin(1/(1/n\pi)) = \sin(n\pi) = 0$. In other words, this sequence is $0,0,0, \dots$. The limit of this sequence is zero.
On the other hand, consider the sequence$$ \frac{1}{\pi/2}, \frac{1}{2\pi + \pi/2}, \frac{1}{4\pi+\pi2}, \dots, \frac{1}{2n\pi + \pi/2}, \dots $$Again, this sequence also converges to $0$. But this time when we evaluate the function at these points, we get $\sin(2n\pi + \pi/2) = \sin(\pi/2) = 1$. So the sequence is just $1,1,1,1, \dots$, and so the limit is $1$ (and not $0$!).
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