Epsilon-Delta Proofs Question

Question

I am somewhat confident that my answers are correct, but would like a second opinion. Below is the question and my reasoning for each:

~Edited per suggestions below.

Question

For which of the functions below is the following statement true:

For any $\epsilon > 0$, there is a $\delta > 0$ such that if $f(x) > \delta$, then $x > \epsilon$.

$$ (1) f(x) = \sin(x)$$ No, this statement does not hold true for $f(x) = \sin(x)$. The sine function oscillates between -1 and 1, and there are values of $x$ for which $x$ is greater than any given positive $\delta$, yet $\sin(x)$ remains within a bounded interval. Therefore, there is no universal $\delta$ satisfying the condition.

$$ (2) g(x) = e^x + 1$$ Yes, this statement holds true for $g(x) = e^x + 1$. For any positive $ \epsilon $, you can choose $ \delta = 1$ such that if $g(x) > \delta $, then $x > \epsilon$.

$$(3) f(x) = -x$$ No, this statement does not hold true for $f(x) = -x$. If $f(x)>\delta>0$ then $x<-\delta<0$ so you cannot get $x>\epsilon$ for any $\epsilon>0.$

$$(4) g(x) = 0$$ No, this statement does not hold true for $g(x) = 0$. The condition $f(x) > \delta$ cannot be satisfied for any positive $\delta$ since $g(x) = 0$ for all $x$. Therefore, there is no $\delta$ that satisfies the condition.

$$(5) f(x) = \begin{cases} x, & \text{if } x > 0 \\ 0, & \text{if } x \leq 0 \end{cases} $$

If $ x > 0$, then $f(x) = x$. In this case, the statement holds true. If you choose $\delta = \epsilon$, then whenever $f(x) = x > \delta = \epsilon$, it implies that $x > \epsilon$. If $x \leq 0$, then $f(x) = 0$. In this case, the condition $f(x) > \delta$ cannot be satisfied for any positive $\delta $ since $f(x) = 0$ for $x \leq 0$. Therefore, there is no $\delta$ that satisfies the condition. Thus, no.

Please let me know if these statements are valid.

Answer 1

(1) Your conclusion is incorrect. You can always take any $\delta>1.$ That will make $f(x)>\delta$ always be false, which implies that the implication "if $f(x)>\delta,$ then $x>\epsilon$" be trivially true. (Note: I first didn't realize the possibility of vaccuous truth in this part.)

(2) Your conclusion is correct, but not your motivation. Firstly, "This function is strictly increasing" doesn't necessarily imply that it's unbounded. For example $\arctan(x)$ is strictly increasing but bounded above by $\pi/2.$ Secondly, your value of $\delta$ for a given $\epsilon$ is not correct.

(3) Incorrect conclusion. If $f(x)>\delta>0$ then $x<-\delta<0$ so you cannot get $x>\epsilon$ for any $\epsilon>0.$

(4) Incorrect conclusion. Nothing says that you must have $f(x)>\delta>0.$ It says "if $f(x)>\delta$". When the condition is always false, then the consequence is always true, no matter what it is.

(5) Incorrect conclusion. Same comment as for (4).

Answer 2

I'm guessing that the purpose of this question is to see if you can work through logic that is vaguely related to the usual type of epsilon-delta proof but is actually quite different. The problem as posed appears to be a useless one, meaning that you're unlikely to have seen it explained anywhere else and need to work through it from first principles. In particular, this is unlike any of the usual epsilon-delta statements you would normally work with, because even though it starts with "for any $\epsilon > 0$, there is a $\delta > 0$" as usual, $\delta$ is compared with $f(x)$ rather than $x$ and $\epsilon$ is compared with $x$ rather than $f(x)$. Also, there are no absolute value signs anywhere.

So you really have to be up to date on your mathematical logic.

The first thing I would recommend is to sear into your memory the fact that in mathematics, $P \implies Q$, which is also written "if $P$ then $Q$," means exactly the same thing as $\lnot P \lor Q$, and therefore $P \implies Q$ is true when $P$ is false. So if you are trying to prove $P \implies Q$, then it is a good thing if you find that $P$ is false in some cases; in fact, if you can prove that $P$ is always false then you have proved that $P \implies Q$ is always true and you don't even have to consider what $Q$ is doing.

Apply this to the statement on the test:

For any $\epsilon > 0$, there is a $\delta > 0$ such that if $f(x) > \delta$, then $x > \epsilon$.

Notice that "if $f(x) > \delta$, then $x > \epsilon$" is true whenever $f(x) > delta$ is false. So whenever you find that $f(x) < \delta$, far from disproving the statement above, you've just confirmed the statement for that value of $x$.

While we're still on the subject of basic mathematical logic, since we're supposed to be able to say whether this statement is true or false given $f$ but not given $x$, we need to quantify $x$ in some way. Presumably, we're meant to quantify $x$ as follows:

$$ \text{For any $\epsilon > 0$, there is a $\delta > 0$ such that for all $x$, if $f(x) > \delta$ then $x > \epsilon$.} \tag A $$

At least, I'm assuming this is how your instructor meant the statement. (I'm also assuming that when your instructor wrote $g(x)$ on the test they meant $f(x)$, and this is not some kind of "Simon says" trick where the answer actually is, "You didn't say $f(x)$.")

Since $P \implies Q$ is equivalent to $\lnot Q \implies \lnot P$ (another useful fact you should burn into your memory), the statement on the test is equivalent to $$ \text{For any $\epsilon > 0$, there is a $\delta > 0$ such that for all $x$, if $x \leq \epsilon$ then $f(x) \leq \delta$.} \tag B $$

You can also write it like this: $$ \text{For any $\epsilon > 0$, there is a $\delta > 0$ such that for all $x \leq \epsilon$, $f(x) \leq \delta$.} \tag {B$'$} $$

Now if $f(x) = 0$, as in part (4) of the question, the statement is true because you can just pick $\delta = 1$ every time and you will always have $f(x) \leq \delta$ whenever you need it. This should be obvious unless you notice that there will always be values of $x$ greater than $\epsilon$, making the $x \leq \epsilon$ clause false, and mistakenly think that that somehow contradicts the statement.

In fact, any function that is bounded above will satisfy the statement. See if you can figure out why.

Perhaps by applying statement (B) to the various examples you will be able to find at least some of the fatal errors in all of the examples except number (3) (in the edit of the question as it was when I wrote this answer).

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Other bits of advice:

Make sure you always remember what role each variable and function is playing in this particular problem. In this problem, if you remember that you have $f(x) > \delta$ in the statement to be proved and not $x > \delta$ or $x < \delta$, then you might realize that the claim "$x$ is greater than any given positive $\delta$" is completely irrelevant to the question.

Make sure you know what the words you use mean and that they actually fit where you use them. The phrase "there is no universal $\delta$" looks completely out of place here. (Is there ever a context where it makes sense?)

Be careful not to let a statement from one sub-problem get used in a different sub-problem where it doesn't belong. Someone said you could just set $\delta = 1$ for one of the parts of the question; then you said $\delta = 1$ in part (2). But if $\delta = 1$ and $e^x + 1 > \delta$, is it really true that $x > \epsilon$ no matter what $\epsilon$ is? Suppose $\epsilon = 1000$: is it true that $e^x + 1 > 1$ implies $x > 1000$?