In the (short) proof for the intermediate value theorem provided in this link: http://www-history.mcs.st-and.ac.uk/~john/analysis/Lectures/L20.html, I didn't understand exactly why $α - δ$ would be an upper bound of $X$ if we assumed $f(α) > 0$. I understand that there would be values of $x$ within $α - δ$ of $α$ for which $f(x)$ would be positive, but how exactly does it qualify the number $α - δ$ as an upper bound for X? Thanks.
Just a doubt on a step in the proof of intermediate value theorem
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calculus
1 Answers
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We have $f(x)>0$ for $\alpha-\delta If there exist $x\in X$ with $x>\alpha-\delta$, then by definition of $X$, we would have $f(y)>0$ for $\alpha-\delta