# Discontinuities definition confusion

Rudin gives the following definition for a right hand limit and left hand limit, and discontinuities (see attached image)

So, to show a function has type 2 discontinuities I need to show that the right hand and left hand limits don't exist.

Should the $$q$$ in the image be an $$x$$?

If not then I am confused because in the definition it already has the right or left hand limit equal to $$q$$ but if I don't know what the left or right hand limit is equal to, how can I then show that $$f(t_{n}) \not \rightarrow q$$ as $$n \rightarrow \infty$$ but $$t_{n} \rightarrow x$$

As an example it has $$f(x) = \left\{\begin{array}{ll} 1 & x\text{ is rational} \\ 0 & x \text{ is irrational} \end{array}\right.$$

It says this has type 2 discontinuities at every point, so the left and right limits do not exist.

So I will take $$\{t_{n}\} = 1 + \frac{\sqrt{2}}{n}$$. Then $$t_{n} \rightarrow 1$$. Then $$f(t_{n}) = 0$$ since $$\{t_{n}\}$$ is irrational for all $$n$$. But I don't see how that doesn't satisfy definition 4.25, since I can just have my $$q = 0$$ and my $$x = 1$$?

• So, to show a function has type 2 discontinuities I need to show that the right hand and left hand limits don't exist. --- This is not quite correct according to Rudin's definition. His definition is a type 2 discontinuity is when you don't have BOTH left and right limits existing, and for this you only need to show that AT LEAST ONE of the left and right limits do not exist (negation of "P AND Q" is "not-P OR not-Q"). Of course, you CAN show a type 2 discontinuity by showing both left and right limits don't exist, but you don't NEED to do this to show a type 2 discontinuity. Sep 20 at 12:59

There are actually two things confusing you.

First:

The $$q$$ in the image is perfectly OK. It is a quantified variable, meaning that it appears in a sentence of the form "If this property about $$q$$ is true, then this other thing is equal to $$q$$."

Imagine a similar sentence: if $$q$$ is the distance from a person's feet to their head, then $$q$$ is the height of that person.

From that sentence, we can conclude:

• if $$q$$ is the distance from my feet to their head, then $$q$$ is my height.
• if $$q$$ is the distance from Lebron James' feet to their head, then $$q$$ is the height of Lebron James.

Both the sentences above are pefrectly true. Yet they do not imply, in any way, that my height is the same as the height of Lebron James.

Second:

You say

So I will take $$\{t_{n}\} = 1 + \frac{\sqrt{2}}{n}$$. Then $$t_{n} \rightarrow 1$$. Then $$f(t_{n}) = 0$$ since $$\{t_{n}\}$$ is irrational for all $$n$$. But I don't see how that doesn't satisfy definition 4.25, since I can just have my $$q = 0$$ and my $$x = 1$$?

The answer to your question is no, you cannot take q. Remember, $$q$$ is the limit if the property listed in the book is true for every sequence $$t_n$$. The property is true for the particular sequence you chose, indeed, but the property is not true for all sequences, as demonstrated by taking the sequence $$t_n = 1+\frac1n$$.

• thanks for this, it makes more sense. So if I want to show that the right or left hand limit does not exist, I need to take two sequences that converge to $x$ and then show that if I apply the function to them I get different values? Sep 20 at 9:50
• @learningmathematics Exactly. Or, you can find one sequence such that when you apply the function to its values, the resulting values do not converge.
– 5xum
Sep 20 at 9:51

In order that $$f(x+)=q$$ we must have $$f(t_n) \to q$$ for every sequence $$t_n$$ decreasing to $$x$$. You are just taking one particular sequence decreasing to $$1$$.

• All great answers and really helpful. On balance will accept 5xum response as it was particularly detailed. Sep 20 at 9:54

To show the right limit exists at $$1$$, for every sequence $$t_n>1$$, $$t_n \to 1$$ we must have $$f(t_n)$$ must be convergent to a single number.

However, you have taken only one sequence and examined it. If you consider another sequence $$s_n = 1 + \frac{1}{n}$$ then $$f(s_n) =1 \to 1$$, so right limit does not exist.