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November 3rd, 2015, 09:12 AM   #1
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Wink Why epsilon and delta can be changed in the definition of continuity at x

We have a function f:[a,b]→R, where [a,b] is an interval bounded by the real straight line. Now someone writes the definition of continuity at x∈[a,b]x∈[a,b] but makes a mistake. He writes δ in stead of ε and ε instead of δ. So the definition changes: for every δ>0 there exists ε>0 such that if x∈[a,b],|x−x0|<δ then |f(x)−f(x0)|<ε

Now we have to prove that this definition is satisfied if and only if the function is bounded by [a,b].

How can I prove this?
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November 3rd, 2015, 09:29 AM   #2
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Suggestion:

Quote:
Now we have to prove that this definition is satisfied if and only if the function is bounded by [a,b].
You don't mean bounded by [a,b], you mean bounded on the closed interval [a,b].
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November 3rd, 2015, 09:37 AM   #3
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Yes sry
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November 3rd, 2015, 09:54 AM   #4
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OK so the usual way to proove if and only if is to start with the contrapositive (imagine it is / is not true and show that leads to a contradiction. Why do you think the boundedness matters, what would happen if were not bounded?
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November 3rd, 2015, 10:33 AM   #5
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For "bounded $\implies$ satisfied", identify a suitable $\epsilon$.
For "satisfied $\implies$ bounded", follow studiot's idea.
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November 3rd, 2015, 12:42 PM   #6
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Can you please be more explicit. I really don't get it.
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November 3rd, 2015, 01:45 PM   #7
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If a function is bounded on $[a,b]$, then there exists an $M \gt 0$ such that $-M \lt f(x) \lt M$ for all $x \in [a,b]$.

If the conditions of the statement are satisfied, then we may pick $x_0 = \frac12(a + b)$ and $\delta \gt \frac12(a - b)$. (Or just pick $\delta \gt b-a$ and any $x_0$). What is the result?
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