# Using arbitrariness of $\varepsilon$ in proofs (metric spaces etc)

Let $(X, d)$ be a metric space. Let $A \subseteq X$ and fix $x \in X$, define $\rho(x, A) = \inf \{d(x,a) : a \in A \}$.

In trying to prove $\overline{A} = \{x \in X : \rho(x, A) = 0\}$, if $x \in \overline{A}$ then for all $\varepsilon>0$, $B_{\varepsilon}(x) \cap A \neq \emptyset$. So let $y_{\varepsilon} \in B_{\varepsilon}(x) \cap A$. By the definition of infimum, we have \begin{align*} \rho(x, A) \le d(x, y_{\varepsilon}) < \varepsilon. \end{align*} Then since $\varepsilon$ was arbitrary, we have $\rho(x, A) = 0$.

What I don't understand is the bolded part, why is it that just because $\varepsilon$ is arbitrary allows us to conclude $\rho(x, A) =0$? What is the fundamental reason behind this? I've seen this kind of argument using the arbitrariness of $\varepsilon$ used elsewhere in really simple arguments, for example:

For all $\varepsilon>0$, assume that $x'+y' > u_1 + u_2 - \varepsilon$ for some real numbers $x', y', u_1, u_2$. Then since $\varepsilon$ was arbitrary, we have $x'+y' \ge u_1 + u_2$. Why? I think I'm missing something really obvious.

Use the following result with $a = \rho(x,A)$ and $b=0$.

Theorem. Let $a$, $b \in \mathbb{R}$. If $a < b + \varepsilon$ for every $\varepsilon > 0$, then $a \leq b$.

(Note that you've already shown the truth of "$a < b + \varepsilon$ for every $\varepsilon > 0$" in your work: when you said that $\varepsilon$ was arbitrary, it means you've shown this inequality to be true for every $\varepsilon > 0$.)

It then follows from the theorem that $\rho(x,A) \leq 0$. But $\rho(x,A) \geq 0$ by definition of $\rho$, so $\rho(x,A) = 0$.

Proof of Theorem. Suppose $a > b$. We show that $a \geq b + \varepsilon$ for some $\varepsilon > 0$. Let $\varepsilon = (a-b)/2$, which is positive since $a > b$. Then $$a = \dfrac{a+a}{2} > \dfrac{a + b}{2} = b + \dfrac{a-b}{2} = b + \varepsilon. \tag*{\blacksquare}$$

Assume that $\rho(x,A)\neq 0$, then $\varepsilon:=\rho(x,A)/2>0$ and by assumption, one has: $$\rho(x,A)<\varepsilon=\rho(x,A)/2.$$ Which leads to $1<1/2$, a contradiction.

The fundamental reason is that if you can prove that some non-negative quantity is smaller than any positive number $\epsilon>0$ then it must be zero, for if it is not zero, then it is some positive number, for which you can always find some smaller $\epsilon>0$.

If you prove that $0\le x<\epsilon$ for an arbitrary $\epsilon>0$ then you prove that $0\le x<\epsilon$ for every $\epsilon>0$.

That is, $x$ is a non negative number that is lesser than any positive number. Since for any positive number there is another lesser positive number (its half, for example), the only non negative number that holds this condition is $0$.

Assume $$\rho(x, A) < \varepsilon$$ for all $\epsilon>0.$ Can be $$\rho(x, A) =a>0?$$ Since $\varepsilon$ is arbitrary we can take $\varepsilon=a/2.$ What do we get in such a case?

$$a=\rho(x, A)<\dfrac a2.$$ This gives us a contradiction. So it must be

$$\rho(x, A) =0.$$

The following answer is a bit cornball, but it is still an earnest attempt to be instructive.

Can the use of '$\text{arbitrary-} \varepsilon$' logic in a proof be an arbitrary choice of techniques? It might even be an indiscriminate choice manifesting abstruseness when other methods are available.

In what follows you will not see any $\varepsilon$'s.

Warm up exercise:

Both the infimum and supremum can be expressed in terms of closure:

Assume the set $K \subset \Bbb R$ has NO greatest element but is bounded from above.

Proposition 1: There is a unique number $k_b \in \overline{K}$ satisfying
$k_b \gt k$ for all $k \in K$.

Proposition 2: The number $k_b$ is the supremum of $K$.

The reader is encouraged to prove both of these statements and the two corresponding propositions for the infimum.

Theorem 3: If $a \gt b$ then there exists a (unique) $p > 0$ so that

$\tag 1 a = b + p$

Proof
If $a \gt b$, then $a - b \gt 0$. Set $p = a - b$. We leave it to the reader to check that (1) holds with this choice of $p$. QED

Corollary 4: If $a \gt b$ then there exists a $q > 0$ so that

$\tag 2 a \ge b + q$

Proof

If you set $q = p/2$ then $a \gt b + q$. But then also $a \ge b + q$, giving (2) . QED

Of course the contrapositive of this corollary is also true:

Theorem 5: Let $a$, $b \in \mathbb{R}$. If $\,a \lt b + q\;$ for every $q \gt 0$, then $a \leq b$.

Observe that it is OK to use a different letter for $q$, like, say, $\text{e-p-s-i-l-o-n}$.

Proposition 6: If $x \in \overline{A}\;$ then $\;\rho(x, A) = 0$.

Proof
Assume $\rho(x, A) = k \gt 0$. It is certainly possible that for some $a \in A$ that $d(x,a) = k$, so we are going to be careful here. The open metric ball about $x$ with radius $k - 10^{-6}$ can't have a nonempty intersection with $A$ (if it did, $\rho(x, A)$ couldn't be a lower bound). So $x \notin \overline{A}$. QED

Exercise: Was it really necessary to be so careful in the above proof?