by Shivam Kumar
Last Updated June 24, 2018 04:20 AM

Let (X,d) be a metric space and let f:X -R and g:X-R be two continuous real valued functions on X , I wish to prove that (f+g) is also continuous on X . My problem is that why we are choosing Delta to be minimum of delta_1 and delta_2 why not maximum of delta_1 and delta_2

By choosing $\delta$ to be the minimum, we have $d(x,y) < \delta$ implies both $d(x,y) < \delta_1$ and $d(x,y) < \delta_2$ so we may use the earlier two continuity statements regarding $f$ and $g$. If you choose the maximum, you may not be able to use one of the earlier statements.

Think about what could happen if $\delta_1\neq \delta_2$. Take $\delta_1>\delta_2$, then if we set $\delta =\text{max}(\delta_1,\delta_2)$ then $\delta=\delta_1$. Thus, we know that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon/2$, but since $d(x,y)<\delta=\delta_1$ and $\delta_1>\delta_2$, we might not have $\delta<\delta_2$, and therefore the second condition might not hold. Therefore, we need to take the minimum of $\delta_1$ and $\delta_2$ to ensure that $d(x,y)<\delta_1$ *and* $d(x,y)<\delta_2$ so that the two implications in the proof hold.

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