Are my examples of False $\implies$ False being true correct? I know that $p \implies q$ is true if $p$ is false. I am trying to wrap my brain around this by thinking about examples. I have two examples that I would like to understand.
Example 1: Consider the set
$$
X = \{x\in \mathbb{R}: x\geq 0 \implies x^2\leq 0\}.
$$
This set is equal to the interval $(-\infty, 0]$ because, for example, $-1 \geq 0 \implies -1\leq 0$ is a true statement. Is that correct?
Example 2: I know that the definition of limit is
$$
\forall \epsilon>0 \exists \delta > 0: 0<\lvert x-a\rvert < \delta \implies \lvert f(x) - L\rvert < \epsilon.
$$
Would this make this a true statement for all functions $f$? Say $\epsilon$ is given, then pick $\delta = -1$. Then $0<\lvert x- a\rvert < \delta$ is false, so it is true that it does imply that $\lvert f(x) - L\rvert < \epsilon$. Have I just misunderstood the definition?
EDIT: I just realized that in my second example the requirement is that $\delta$ be positive. I guess I am down to the first example.
 A: Some people make a distinction between $\implies$ implication and $\rightarrow$ implication. But if you're treating $\implies$ as the simple kind of material conditional that might appear in a truth table, as I think you are,
then your first example is correct, because the sentence
$$ x\geq 0 \implies x^2\leq 0 $$
(with free variable $x$) is equivalent to the sentence
$$ x \leq 0. $$
In this example, because of the use of the set-builder notation
$\{x\in \mathbb{R}:\cdots\}$, there is implicitly a quantification that says $x$ must be in $\mathbb R$ but does not force $x$ to be in any particular proper subset of $\mathbb R$. Only the implication removes the positive values of $x.$
The set-builder notation is also what tells us that
$x\geq 0 \implies x^2\leq 0$ is not universally quantified,
that is, we are not being asked whether there are any counterexamples.
We're merely being told that the counterexamples (if they exist) don't belong to the set that is being constructed.

Edit: I see you figured out the following on your own just before I posted, but here is what I had to say about the original version of the question:
In example 2, however, the implication is quantified by $\exists \delta > 0$.
This is what prevents you from supposing the implication could be true by virtue of setting $\delta = -1.$
A: Be careful about mixing a general mathematical statement with a specific instance.  There are typically two ways to interpret $\Rightarrow$.
As an operator: This is thinking of $\Rightarrow$ as a function on a pair of Boolean values.  In Example 1, you're thinking of $\{x\in\mathbb{R}:x\geq 0\Rightarrow x^2\leq0\}$ as the set of all real $x$ such that for fixed $x$, the statement $x\geq 0\Rightarrow x^2\leq 0$ is true.  This is where $(-\infty,0]$ comes from.  In David K.'s answer, this would correspond to $x\geq 0\rightarrow x^2\leq 0$.  I think that using this notation would make this example much clearer.
As a mathematical statement: Typically, when we write $p\Rightarrow q$ in math, we're thinking about a universally quantified statement.  In other words, we say that a statement is true if for all possible instances, the corresponding implication is true.  In this case, the statement $x\geq 0\Rightarrow x^2\leq 0$ is false since there exist counterexamples, i.e., when $x=1$, the hypothesis is true, but the conclusion is false.  Therefore, as a universally quantified statement, this is false.  In this case, the set would be empty.  This may not be what is intended in example since the same $x$ is used in both places in the set-builder notation.
Overall, I think that your Example 1 is correct, but the use of $\Rightarrow$ can be quite confusing when used in this way.  I would write this as $\{x\in\mathbb{R}:(x<0)\vee(x^2\leq 0)\}$ to be absolutely clear about what you're intending.
A: For your Example 1, for $x\in \mathbb{R}$ let $S(x)$ be the statement $[x\ge 0 \implies x^2\le 0]$. The set can be written as $X=\left\{x\in \mathbb{R}: S(x)\,\text{is true}\right\}$. Since $-1\ge 0$ is false, the statement $S(-1)=[-1\ge 0\implies (-1)^2\le 0]$ is true. Therefore, I do believe that you are right.
The truth behind logical implication seems to be subtle. I have always accepted truth behind the statement $p\implies q$ as a matter of definition, the definition coming from the truth table for $p\implies q$. My guess is that a logician will give a better answer to this.
This all said, I do think you can develop intuition as to why false $\implies$ false should be true. For example, consider the statement "If you score a touchdown, I will buy you a gift." You can write this as:
$$\text{You score touchdown}\implies\text{I buy you a gift}.$$
If you end up not scoring a touchdown, and, as a result I do not buy you a gift, then both statements above are false, and therefore the implication is true. The fact of the matter is that I did not break my promise of buying you a gift, because you did not score a touchdown for me! So nothing "false" occurred. On the other hand, if you had scored the touchdown, and I didn't buy you a gift, then I broke my promise. In this case it was a "false promise." Although advanced mathematics is more abstract than real world settings like these, I like to think about such examples to understand $p\implies q$ more generally.
A: In example 1, rather than reading $X = \{x\in \mathbb{R}: x\geq 0 \implies x^2\leq 0\}$ as if it says that $X$ is the set of non-negative reals whose squares are less than or equal to $0$, you are reading it as the set of reals which if non-negative have squares less than or equal to $0$.  So instead of $X=\{0\}$ you end up with the non-positive reals and this is easy to demonstrate:

It is a property of an ordered field that if $a \le 0$ and $b \ge 0$ then $ab \le 0$.
So, for any $a \le 0$, you have $a \ge 0 \implies a^2 \le 0$

and this might be justified as a literal if unintended reading
For the limit of $f(x)$ being $L$ as $x \to a$, I think a better definition might be $$\forall \epsilon>0 \,\exists \delta > 0\, \forall x\in \mathbb R: 0<\lvert x-a\rvert < \delta \implies \lvert f(x) - L\rvert < \epsilon$$ which would not allow your reading
A: Your first example is technically correct. I say technically because there seems to be a subtle difference between the actual definition of $\implies$ and the way most mathematicians use it. Of course, $x\geq 0 \implies x^2 \leq 0$ is a well-formed formula, and by the axiom of specification it follows that $X$ exists. What is in it? Elements of $\mathbb{R}$ that satisfy the above implication. And these are, as you correctly pointed out, $(-\infty, 0]$. Call this the axiomatic interpretation if you like. For most mathematicians, however, material implication isn't a definitive truth function, but rather a common tool, a shorthand for saying that if the antecedent is correct, then the consequent is, too. As for the other truth configurations, they just don't bother, intuitively brushing them off as "undefined" or simply irrelevant (note that this usually doesn't prevent them from reaping the mathematical power of contradiction proofs, contrapositives etc.) Whenever used axiomatically or in logical contexts, you'd often find that the skinny arrow $\to$ notation is preferred, whereas the common usage seems to rely on $\implies$.
For your second example, things are not quite so straightforward. Here, we have to prove the implication holds for all $x$ in the domain of $f$. The last part is often suppressed in the $\varepsilon$-$\delta$ limit definition, but it is there. Observe that the only thing we actually need to prove is that whenever the antecedent is true, the consequent is true as well (if the antecedent is false, we don't care either way, as the implication is satisfied). Note, however, that now we have quantifiers bounding some of the variables, and for us to claim our proof as successful, the valuation of the overall expression has to be $\mathbb{T}$, not just the implication bit. For instance, this is precisely why your example of $\delta = -1$ is false: $\delta$ is bounded by $\forall \delta > 0$. But suppose you pick another $\delta$, positive this time. And this is where the hidden bit $\forall x\in \text{Domain}(f)$ comes in. No matter how small $\delta$ is, we'll always be able to find $x$ close enough to $a$ to make $0< \lvert x-a\rvert <\delta$ true, and unless you had a very lucky choice of $\delta$, you'd end up with $\mathbb{T}\implies \mathbb{F}$, i.e. a false implication.
A: Yes, the statement in your first example is indeed a true statement.
Note that when $p$ is false, the constraint is essentially removed from $q$, therefore we can't really say much about $p\rightarrow q$. However, $p$ being false doesn't necessarily contradict the material implication either. This is why, by convention, $p\rightarrow q$ is true when $p$ is false. I believe this gets to the heart of the matter.
