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I came across this theorem and its proof while studying Real Analysis on my own. I don't understand how does the contradiction as claimed in the second last line hold when $d$ is less than or equal to $0$. If anybody could shed some light on this, it would be very helpful.

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  • $\begingroup$ You are right. You need to take $0 < d < \min\ \{\delta, c-a \},$ which is possible. The author has implicitly assumed that $d$ is a positive quantity. $\endgroup$
    – Anil Bagchi.
    Sep 6, 2020 at 5:41

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It isn’t explicitly stated (though it probably should be), but $d$ is chosen to be positive. This is possible because $\delta>0$, and it’s been proved that $a<c$, so that $c-a>0$. Thus, $\min\{\delta,c-a\}>0$, and we can choose $d$ so that $0<d<\min\{\delta,c-a\}$.

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