Continuous function
A useful lemma

Let f (x) {\displaystyle f(x)} be a function that is continuous at a point x 0, {\displaystyle x_{0},} and y 0 {\displaystyle y_{0}} be a value such f (x 0) ≠ y 0. {\displaystyle f\left(x_{0}\right)\neq y_{0}.} Then f (x) ≠ y 0 {\displaystyle f(x)\neq y_{0}} throughout some neighbourhood of x 0. {\displaystyle x_{0}.}

Proof: By the definition of continuity, take ε = | y 0 − f (x 0) | 2 > 0 {\displaystyle \varepsilon ={\frac {|y_{0}-f(x_{0})|}{2}}>0}, then there exists δ > 0 {\displaystyle \delta >0} such that | f (x) − f (x 0) | < | y 0 − f (x 0) | 2 whenever | x − x 0 | < δ {\displaystyle \left|f(x)-f(x_{0})\right|<{\frac {\left|y_{0}-f(x_{0})\right|}{2}}\quad {\text{ whenever }}\quad |x-x_{0}|<\delta } Suppose there is a point in the neighbourhood | x − x 0 | < δ {\displaystyle |x-x_{0}|<\delta } for which f (x) = y 0; {\displaystyle f(x)=y_{0};} then we have the contradiction | f (x 0) − y 0 | < | f (x 0) − y 0 | 2. {\displaystyle \left|f(x_{0})-y_{0}\right|<{\frac {\left|f(x_{0})-y_{0}\right|}{2}}.}
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