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My professor has a way of doing proofs, for example, when proving that the Darboux integral and reiman have the same value, she did something like (having assumed darboux integrability):

$r - \epsilon < L(f) \le U(f) < r + \epsilon$

and said let $\epsilon \to 0$ and we have $r = U(f) = L(f)$.

Why is this mathematically and logically sound and correct to do?

Here's my rough understanding:

  • we have this for all $\epsilon$.
  • letting it go to 0 gets us close to it, but never equal, in a sense, like a limit.

She does not do this by definition and I'd never seen this strategy before, so I am wondering about it.

Math solver
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user129393192
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2 Answers2

5

Suppose $b<r+\varepsilon$ holds for all $\varepsilon>0$. We claim that $b \leq r$. Indeed, if $b>r$, then $b-r>0$ and we know that $$ b<r+(b-r) = b, $$ which is absurd. Hence, $b \leq r$.

Similarly, if $r-\varepsilon<a$ for all $\varepsilon>0$, then $r \leq a$.

Thus, if $r-\varepsilon<a \leq b<r+\varepsilon$ for all $\varepsilon>0$, then $r \leq a \leq b \leq r$, so that $a$ and $b$ are both equal to $r$.

azif00
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3

I was able to figure it out. Essentially, we'd like to say that if $|x - y| < \epsilon$ for all $\epsilon > 0$ then $x = y$

Proof:

Suppose $x \ne y$

then $|x - y| > 0$

let $\epsilon = |x - y|$

then $|x - y| < |x - y|$. Contradiction.

thus $x = y$

user129393192
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