Difference between two statements/quantifiers?

Question

1) $\forall \varepsilon > 0$, if $a + \varepsilon > b$, then $a > b$.

2) If $\forall \varepsilon > 0, a + \varepsilon > b$, then $a > b$.

When I read these two statements, they seem to imply the exact same thing, but apparently they mean two different things. What is the difference?

Answer 1

The difference lies in what it takes to prove or disprove the statements.

In the first example, to prove that the statement is false you need merely find one value of $\varepsilon$ such that $\varepsilon > 0$ is true but $a + \varepsilon > b \Rightarrow a > b$ is false. In the second example, to prove the statement is true it is sufficient that for any $a$ and $b$ you can show either that $a > b$ or that there is some $\varepsilon$ such that $\varepsilon > 0$ is true but $a + \varepsilon > b$ is false.

One thing that is not shown in these somewhat informal statements is how you quantify $a$ and $b$. Since these are not formally quantified, we interpret these statements as asserting something for all values of $a$ and $b$. We can also be a lot more explicit about exactly what is the "if" part of the implication and what is the "then" part of the implication by writing the implication in the form $P \Rightarrow Q$ with parentheses to show exactly where $P$ begins and ends.

1. $\forall a\colon \forall b\colon \forall \varepsilon > 0\colon (a + \varepsilon > b \Rightarrow a > b)$.

2. $\forall a\colon \forall b\colon (\forall \varepsilon > 0\colon a + \varepsilon > b) \Rightarrow a > b$.

It is easy to find a counterexample for the first statement; for example, let $a = 1$, $b = 2$, and $\varepsilon = 9$.

We can use rules of formal logic to rewrite the second statement as follows:

$$\forall a\colon \forall b\colon (\neg(\forall \varepsilon > 0\colon a + \varepsilon > b) \vee (a > b)).$$ $$\forall a\colon \forall b\colon ((a \leq b) \Rightarrow \neg(\forall \varepsilon > 0\colon a + \varepsilon > b).$$ $$\forall a\colon \forall b\colon ((a \leq b) \Rightarrow (\exists \varepsilon > 0\colon a + \varepsilon \leq b).$$

We can now use the example $a = b = 1$ to prove the statement is false: $1 \leq 1$, but $\forall \varepsilon > 0\colon 1 + \varepsilon > 1$.

But suppose we change the statements very slightly:

3. $\forall a\colon \forall b\colon \forall \varepsilon > 0\colon (a + \varepsilon > b \Rightarrow a \geq b)$.

4. $\forall a\colon \forall b\colon (\forall \varepsilon > 0\colon a + \varepsilon > b) \Rightarrow a \geq b$.

Statement 3 is (still) false, but statement 4 is true, because if $a < b$ then there is some $\varepsilon > 0$ such that $a + \varepsilon \leq b$.

Answer 2

What's being said in $1)$ is that: taken any positive real number $\epsilon$, if we have two real numbers $a$ and $b$ for which: $a + \epsilon > b$, then we would have $a > b$. Here, the choice of $\epsilon$ is prior to the choice of $a$ and $b$. The statement $a + \epsilon > b$ here does not hold for any positive $\epsilon$, it just holds for the one we have already chosen before.

While $2)$ says that: If we have two real numbers $a$ and $b$, such that taken any positive real number $\epsilon$, we have $a + \epsilon > b$, then $a > b$. Here, we already have the two real numbers $a$ and $b$, and the statement $a + \epsilon > b$ holds for any $\epsilon > 0$ we would like to choose.