Epsilon delta definition of continuous functions

A soft 101-introduction

Remember my rule: you aren't allowed to read this if it doesn't interest you. Well, you can start reading but then if it bores you you gotta stop! I only do this for fun and because I love it and so should you.

I'm going to teach you a lesson taught in university math. Even if you’re just starting high school, I’ll bet that if you try hard, you’ll be able to understand this. This lesson usually causes a lot of headaches for students, but keep reading, and I think you’ll be fine.

For everyone who has no idea wtf a continuous function is, it’s a function that is relatively unbroken. Very loosely speaking, here is a continuous function and a non continuous one:

Pic 1: This red graph is a continuous function

Pic 2: This blue graph is NOT a continuous function

See that “break” in the blue curve? That’s the problem spot. That’s the point at which this function is not continuous. At every other \(x\) value, the blue function is continuous. 

The high school definition of a continuous function is essentially, “A function is continuous if you can draw it without taking your pen off the paper”. This definition is OK for intuition, but in university you’re going to need to understand some language and symbols and concepts that are more precise than this. 

That’s what this post is about. It’s about getting a taste for university precision regarding the definition of continuity. 

Here’s the definition (broken into easy language, but still retaining most of its precision) of a continuous function at a point \(x\) - which I want you to READ but you are NOT expected to understand it. Just see the words, and store them in your brain so that you are familiar with them as we go:

A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\) (on the \(y\)-axis), we can find a related interval around \(x\) (on the \(x\)-axis) where the images of any point in that interval are completely within the initial interval we picked around the image of \(x\).

Woah. Our goal in this post is to understand THAT. Fully. Completely. Now get pumped up!

Let’s talk about what a few of those words mean. That’s always the first step into understanding a statement: figure out exactly what all the words mean.
Let’s talk about intervals, and image. Those are really the only two terms that might be new.

An interval around a point is just a group of points that are within a certain distance. For example, you might say that a \(5\)km radius from your house is a \(5\)km interval (note that this would then be a \(10\)km diameter). Another example of an interval would be the interval of radius \(1\) on the real line, centred around the point \(x=5\). That would look like this:

Pic 3: Interval on the real line of radius \(1\) around the point \(x=5\).

See how \(5\) is in the centre, and how every point on the red line is at most a distance of \(1\) away from \(5\)? That my friends, is an interval of radius \(1\) around the \(x\) value \(5\) (assuming we are restricted to the real line).The image of a point \(x\) is the point that our function sends \(x\) to. The image then, is a \(y\) value. It’s a point on the \(y\)-axis. If \(x\) is the input, \(f(x)\) is the image, or output. In the usual case of having a function that sends a real number to another real number, the image of a point \(x\) is just the height of the function at that particular \(x\) value. I like to think of it as the length of the shadow of \(x\).

Pic 4: The length of the blue line represents the image of \(x=5\)

Here is the definition broken into several chunks, which we’ll build on until we have the full definition. I’ll explain each number separately so it’ll be easier to understand.

  1. A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\)
  2. A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\), we can find a related interval around \(x\)
  3. A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\), we can find a related interval around \(x\) where the images of every point in that interval fall completely within the initial interval we picked around the image of \(x\).

Let's start from the top.

  1. A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\)

First thing to note is that for now, we are not asking if the whole function is continuous. We are only interested in seeing if a function is continuous at a single point \(x\). So the very first thing we do is pick the \(x\) value we’d like to analyse.

Now, we know what the image of \(x\) is (it’s the length of that blue line in Pic 4, directly above). Now what is an interval around the image? In this case, it’s important to note that an interval around the image is referring to an interval on the \(y\)-axis.

For example,

Pic 5: The green lines form an interval around the image of \(x=5\).

In this graph, the horizontal orange line represents the height of the image of \(x=5\). The two green lines are each a small distance away, creating a small interval around the image.

Important question for later: Can you see the different \(x\) values that produce images that fall within the two green lines?? Think about this until you know the answer or are completely stumped. The answer is below so don't read it until you've tried to figure it out. Invest one or two minutes.

Answer only after you've thought about it: It looks like the image of \(x=3\) would definitely be in the range. Same with \(x=5.1\), or \(4.7\). There are many many many \(x\) values whose images fall within the accepted range. In fact, Here is a picture showing all the \(x\)-values whose images fall within the desired range. Any \(x\)-value that’s part of a purple column will do the trick.

Pic 6: The purple columns are there to help you visualize which \(x\) values have images that fall within the accepted green lines. Note that all \(x\) values that are not part of a purple column will have an image that is not between the green lines.

It's important to note that for this particular green interval around the image of \(x=5\), we CAN find an interval around \(5\) whose images all fall within the desired range.

What happens if we pick a smaller interval around the image of \(x=5\)? Let’s take a look:

Pic 7: the green lines are tighter due to a smaller interval around the image of \(x\). As a result, we now have fewer accepted \(x\) values whose image falls within the green lines.

With a smaller interval, the green lines hug the orange line a little closer. As a consequence, you’ll notice that there are far fewer \(x\) values whose images fall within the new, smaller interval. since fewer \(x\) values now have images that fall within the green lines, the purple columns will get thinner as the green lines get tighter.

OK so now we know what an interval around the image is, and we know that if you make the size of your interval around the image smaller, your purple column gets smaller as there are now fewer acceptable \(x\) values.

2. For ALL intervals around the image of \(x\), we can find a corresponding interval around \(x\)

If you understood point \(1\), this one will be a cinch. The “interval around \(x\)” are all the \(x\) values in the purple column. The only new concept here is that we’re finding a corresponding interval around \(x\), which sits somewhere on the \(x\)-axis. In other words, our interval around the image of \(x\) is on the \(y\)-axis, while this interval is on the \(x\)-axis.

3. For ALL intervals around the image of \(x\), we can find a corresponding interval around \(x\) where all their images fall within the interval around \(x\)’s image.

Note that every \(x\) value in the purple column has an image that fits between the green lines, since that’s the way we chose the purple column.

The essence of a function to be continuous at a point \(x\), is that we must be able to guarantee that for ANY interval around the image of \(x\) (on the \(y\)-axis), we can find a corresponding interval around \(x\) (on the \(x\)-axis) such that any of the images of the points in the \(x\) interval fall within the interval on the \(y\)-axis.

If our function \(f\) is continuous at \(x\), it means that no matter how tightly you make the green lines hug the orange line, you will always be able to find a corresponding purple column, with accepted \(x\) values on BOTH sides of your chosen \(x\) value (that's part of what an interval AROUND something means). If you can always do this around the point \(x=5\), we say that the function \(f\) (the red line) is continuous at the point \(x=5\).

That’s the language portion of this post! Now onto the logic!

Pic 8: A clearly non-continuous function at \(x=2\) 

Clearly, this function is NOT continuous at \(x=2\). There is a jump there which fucks us over, hard. But that’s not the interesting part. The interesting part is determining WHY this function fails the definition of continuity that we just took the time to understand.

This function fails the test for continuity because there are some intervals around the image of \(2\) that do NOT have a corresponding interval around \(2\) with the required property that those \(x\) values all have images that fall within the interval around the image. Man, that is such a mouthful.

To help understand, go here. Play with the \(a\) value, which in this case is acting like our epsilon (from this point forward, we will use the Greek letter \(\epsilon\) instead of the word epsilon) . Note that for \(\epsilon\) values less than \(1\), we can find no interval around \(2\) on the \(x\) axis that meets the requirements given in the definition. Thus, Not continuous at \(x=2\).

Now, if you understand everything up to here, then you actually already understand what it means for a function to be continuous according to the definition you’d see in university. If you feel like taking it one step further, which I certainly do, then let’s try and understand the language that is typically used in more advanced textbooks.

Let’s start using the following version of the definition, which means the EXACT same thing, except it’s a little more precise, and has way fewer words.

“For any \(\epsilon>0\), we can find a \(\delta>0\) such that \(|f(x)-f(a)|<\epsilon\) whenever \(|x-a|<\delta\).”

If you need to brush up on your absolute values, go ahead and do that. I won’t talk too much about them here. I will say this: If you write \(|x-2|<3\), for example, it represents an interval on the \(x\) axis. Which interval? It’s all the \(x\) values that are a distance of less than \(3\) away from \(2\).

If you don’t believe me, which you shouldn’t unless you actually SEE what’s going on, you should plug a few numbers into \(|x-2|<3\) to see which numbers satisfy the condition and which numbers do not.

Let’s make \(x=1\). Does it satisfy \(|x-2|<3\)? Check: \(|1-2|=|-1|=1<3\), so yes, \(x=1\) is a solution.

Using similar logic, you’ll see that the only \(x\) values that work are the \(x\) values that are less than a distance of \(3\) away from the number \(2\).

If we then replace the \(2\) with an \(a\), like in \(|x-a|<3\), we’re now talking about all the \(x\) values that are a distance of less than \(3\) away from an unknown value \(a\). So, as \(a\) moves around the \(x\)-axis, our set of solutions will also move around to hug whatever value we assign to \(a\).

Finally, if we write \(|x-a|<\delta\), it still represents all the \(x\) values that are less than a certain distance away from \(a\), it’s just that we haven’t specified our distance, and we leave it as the value \(\delta\) (the greek letter delta).

Now, the other part of the new definition is \(|f(x)-f(a)|<\epsilon\). This will work the same way as our interval on the \(x\)-axis, but now we’re on the \(y\) axis. How can I tell? I can tell because we are using \(f(x)\), and \(f(a)\), which are the images of \(x\) and \(a\), respectively. Images are always represented by values on the \(y\)-axis.

As you might then expect, \(|f(x)-f(a)|\) represents the distance that \(f(x)\) is away from \(f(a)\). In our definition, we want it to be less than an unknown value, \(\epsilon\).

OK so let’s compare the two definitions we’ve used thus far, and try and see if we can see that they are really the same thing

Kinda shitty definition: A function \(f\) is continuous at a point \(x\) if for any small interval around the image of \(x\) (on the \(y\)-axis), we can find a related interval around \(x\) (on the \(x\)-axis) where the images of any point in that interval are completely within the initial interval we picked around the image of \(x\).

Much better def, which is different in language but IDENTICAL in content: For any \(\epsilon>0\), we can find a \(\delta>0\) such that \(|f(x)-f(a)|<\epsilon\) whenever \(|x-a|<\delta\).

Can you believe that’s the whole definition? Crazy right? You can see why mathematicians take the time to introduce complicated symbols. If they didn’t, it would take PAGES to explain concepts that you could otherwise explain in a sentence.

Here's an example to help drive home the concepts.

Pic 9: A continuous function given by the line \(y=x+1\) 

  1. At which \(x\) values is the function continuous?
  2. For any \(\epsilon\) value, which \(\delta\) value will guarantee the continuity condition is satisfied?

Answers:

1: This function is continuous at every \(x\) value.

2: This line has a slope of \(1\). That means that if we take an \(\epsilon\) interval around the image of any point, that top line of the interval will intersect our function exactly \(\epsilon\)  units to the right (slope of \(1/1\) means the line always has the same rise and run), and our bottom line of the interval will intersect our function exactly \(\epsilon\) units to the left. So, why not make \(\delta=\frac{\epsilon}{2}\) (anything less than \(\epsilon\) will do the trick)? That way we guarantee that we’ll always be within the required range.

To help see this, check this graph out. Notice that no matter how small you make your \(a\) value, there is always an interval around \(2\) that produce images within \(a\) units of \(f(2)\).

OK, if you made it this far you totally rock! Like, really, you do! I had a really fun time making this doc. Please send any comments you like, and yes I’ll read every single one. If you weren’t able to follow this whole thing, please tell me where it got confusing!

HTML Comment Box is loading comments...