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I'm having trouble with problem 1-19 in Spivak's Calculus. I have to prove that if $|x-x_0| < \frac{\epsilon}{2} $ and $ |y-y_0| < \frac{\epsilon}{2} $ then $ |(x-y)-(x_0-y_0)| < \epsilon $. I know $ |x-x_0| - |y-y_0| < \epsilon$, but since $|a| - |b| \leq |a-b|$, my guess is that this is useless?

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  • $\begingroup$ look up triangle inequality. $\endgroup$ – abel Jan 25 '15 at 16:43
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You can use the triangle inequality like this: $$|(x-y)-(x_0-y_0)|=|(x-x_0)+(y_0-y)|\le|x-x_0|+|y_0-y|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$$

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I know I'm super late, but regardless I'd like to give some input that may help anyone who comes across this question in the future.

you wrote:

I know $ |x-x_0| - |y-y_0| < \varepsilon$, but since $|a| - |b| \leq |a-b|$, my guess is that this is useless?

but that's not correct! In fact that's an important remark to make (albeit an ultimately unnecessary one) because it provides some connection between the two in your brain. But as you'll see you don't need to even need to technically make that statement.

Using your notation where $a = (x-x_0)$ and $b = (y-y_0)$, we want to prove that

$ |(x-y)-(x_0-y_0)| = |a - b| < \varepsilon$

Since

$|a| < \frac{\varepsilon}{2} $ and $|b| < \frac{\varepsilon}{2} $

then

$|a| + |b| < \varepsilon$

And we can show that $|a-b| \leq |a| + |b|$ (by comparing the squares of the two sides of the inequality). And since we've already made the statement that $|a| + |b| < \varepsilon$, we've now proven that $ |a - b| = |(x-y)-(x_0-y_0)| < \varepsilon$.

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You are close:

$|(x-y)-(x_0-y_0)|=|(x-x_0)+(-y+y_0)| \leq|x-x_0|+|y-y_0|$

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