Intuition about : $a = b \iff | a − b| < \epsilon$, for every $\epsilon > 0$

Question

This is Theorem 1.2.6 in Abbott. Understanding Analysis (2016 2 edn). pp 9 - 10. I'm NOT asking about proof that the author proves. Please don't prove. I'm longing just for intuition.

1. I can't intuit how, God willing, an equality (on LHS) can be equivalent to a conjunction of strict inequalities (on RHS).

2. Is there any picture that can assist?

Answer 1

Identity of Indiscernibles :

If we cannot find the slightest difference between two objects $a$ and $b$, then necessarily the two must be identical.

And vice versa.

Answer 2

Suppose $|a-b| < \epsilon$ for all $\epsilon > 0$. Thus the distance between $a$ and $b$ is smaller than every strictly positive number.

Suppose to the contrary that this distance is non-zero, then $|a-b| > 0$ and then your claim says that $|a-b| < |a-b|$, which is a contradiction.

Thus we must have that the distance $|a-b| = 0$, which only happens when $a=b$.

Drawing a picture to see what's going on will definitely help. Take two distinct points $a\neq b$. On the picture, argue why $|a-b| < \epsilon$ for all $\epsilon > 0$ can never happen (my proof above contains the answer). So I would argue that the proof is actually the intuition.

Answer 3

For $x$ non-negative, $\forall \epsilon>0:x<\epsilon$ is a contrived way to say $x=0$ without saying it.

Because if $x$ is positive, you will find some $\epsilon$ that contradicts the inequality $x<\epsilon$ (for example $\epsilon=\frac x2$).

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Note that $\forall \epsilon>0:x\le\epsilon$ also works. But not $\forall \epsilon\ge0:x<\epsilon$.

Answer 4

Short answer:

Smaller than anything can only be zero.