$|x-a| < \epsilon$ , $|y-a| < \epsilon$ (where $x,y,a\in\mathbb R $ and $\epsilon>0$). Use the Triangle Inequality to find an estimate for $|x-y|$

Question

I am teaching myself Real Analysis. I have done proofs in Discrete Mathematics, Linear Algebra, and Abstract Algebra. Nothing too difficult, mainly elementary and intermediate kinds! For this problem, I will lay out my thought process and explain where I am stuck at. If someone could please explain, in a clear manner, what to do OR giving me a hint, that would be helpful!

My thought:

Consider $|x-a| < \epsilon$. If we add $-|y-a|$ to both sides of the inequality, then the inequality maintains order and we have $|x-a| - |y-a| < \epsilon -|y-a|$. By the Triangle Inequality, we can see that $|x-a| - |y-a| \leq |x-y|$.

Now this is where I am stuck. I am not quite sure on how to explain if $|x-y| < \epsilon - |y-a|$, $|x-y| \leq \epsilon - |y-a|$, or if there is something else I am missing.

Any help explained clearly would be kindly appreciated.

Answer 1

Using the triangle inequality,

$|x-y|=|(x-a)+(a-y)|\le|x-a|+|a-y|=|x-a|+|y-a|\lt \epsilon + \epsilon=2\epsilon$.

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Alternatively $|x-a|<\epsilon$ and $|y-a|<\epsilon\iff a-\epsilon<x<a+\epsilon$ and $a-\epsilon<y<a+\epsilon\iff$

$ a-\epsilon<x<a+\epsilon$ and $-a-\epsilon<-y<-a+\epsilon\implies-2\epsilon<x-y<2\epsilon\iff|x-y|<2\epsilon$.

Answer 2

$|x-y| = |(x-a) + (a-y)| \le |x-a| + |a-y| < 2\epsilon$.

That's the easy part. You did the hard part that I, personally, always screw up.

Actually ... you say "we have |x−a|−|y−a|<ϵ−|y−a|. By the Triangle Inequality, we can see that |x−a|−|y−a|≤|x−y|". Could you explain why? Is $|x-y| \ge \epsilon -|y-a|$? why do you have strict inequality? How can you justify that?

Let me try and let's hope I don't screw up.

$|x-a|= |(x-y)+(y-a)| \le |x-y|+|y-a|$

$|x-a|-|y-a| \le |x-y|$.

Of course if $|x-a| < |y-a|$ the left hand side is negative and the result is trivial but we could just as easily proven $|y-a| = |(y-x) +(x-a)| \le |x-y| -|x-a|$ and concluded $|y-a| -|x-a| \le |x-y|$.

So $0\le ||x-a| -|y-a|| \le |x-y| < 2\epsilon$.