Explain how to take the logical negation of the precise definition of continuity.

Question

1. Explain what this statement means in plain English.
2. I am really struggling with understanding this topic

Answer 1

To negate a statement means to make it say the opposite thing. For example, a negation of "it is either raining or cloudy" would be "it is not raining and it is not cloudy." So to negate the precise definition of continuity, first start by stating it. There are several equivalent but very different-sounding statements so I'm not sure which one you wanted to negate, but one of the most common ones to see in calc class is what we call the "epsilon-delta definition."

Statement:

The function f is continuous at a point c if for all 𝜖 >0, there exists 𝛿>0 such that if |𝑥−c|<𝛿, then |𝑓(𝑥)−𝑓(c)|<𝜖.

Intuition of that statement:

Let's focus in on the "for all 𝜖 >0, there exists 𝛿>0 such that whenever |𝑥−c|<𝛿, we have |𝑓(𝑥)−𝑓(c)|<𝜖" part. This is saying, no matter what 𝜖 you pick, you'll be able to find a 𝛿 such that if you take a point in the domain that's within 𝛿 of c, the function's value at that point is within 𝜖 of f(c). In plainer English, this is saying, if you look at the graph of the function and you want to find a point on the curve that is super close to the point (c, f(c)), then you can - no matter how close "super close" is. This makes intuitive sense since it's just saying that there are no breaks or jumps in that curve, so you can move close to c on the x-axis and end up close to f(c) on the y-axis.

Negation:

Okay so now how do we negate this? Let's negate the intuitive statement first, and then work on the mathematical one. The intuitive statement says

if you look at the graph of the function and you want to find a point on the curve that is super close to the point (c, f(c)), then you can - no matter how close "super close" is.

To negate that, let's try to get a statement that says the opposite thing:

there's a point c in the domain of the function such that no matter what point x you pick in the domain, (x,f(x)) is not going to be as close as you wanted it to be to (c,f(c))

Okay so that's what the negation should mean, so now let's try to negate the formal statement. I'll underline the things that I changed from the original statement while negating:

The function f fails to be continuous at a point c if there exists 𝜖 >0 such that for all 𝛿>0, we have |𝑥−c|<𝛿 and |𝑓(𝑥)−𝑓(c)|≥𝜖.

Notice that where it said "there exists" now it says "for all," and vice versa!