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Note that, due to the continuity of <math>f</math> at <math>a</math>, we can keep <math>f(x)</math> within any <math>\varepsilon>0</math> of <math>f(a)</math> by keeping <math>x</math> sufficiently close to <math>a</math>. Since <math>f(a)<u</math> is a strict inequality, consider the implication when <math>\varepsilon</math> is the distance between <math>u</math> and <math>f(a)</math>. No <math>x</math> sufficiently close to <math>a</math> can then make <math>f(x)</math> greater than or equal to <math>u</math>, which means there are values greater than <math>a</math> in <math>S</math>. A more detailed proof goes like this:
Choose <math>\varepsilon=u-f(a)>0</math>. Then <math>\exists \delta>0</math> such that <math>\forall x \in [a,b]</math>, <math display="block">|x-a|<\delta \implies |f(x)-f(a)|<u-f(a) \implies f(x)<u.</math>Consider the interval <math>[a,\min(a+\delta,b))=I_1</math>. Notice that <math>I_1 \subseteq [a,b]</math> and every <math>x \in I_1</math> satisfies the condition <math>|x-a|<\delta</math>. Therefore for every <math>x \in I_1</math> we have <math>f(x)<u</math>. Hence <math>c</math> cannot be <math>a</math>.
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