Let $f:I\rightarrow R$ be continuous at $y\in I$. Suppose $f(y)>m$ for some $m \in R$. Prove there exists $\delta >0$ such that $f(x)>m$ for all $x \in I$ with $|x−y|<\delta $.

Proof: Let $f:I \rightarrow R$ be continuous at $y \in I$. By definition of continuity, $\forall \varepsilon >0 \, \exists \delta > 0$ such that if $x\in I$ and $|x−y|<\delta $ then $|f(x)−f(y)|<\varepsilon$. Suppose $f(y)>m$ for some $m\in R$. And that's where I get stuck.


Let $\epsilon<f(y)-m$. Now apply the definition of continuity.

  • $\begingroup$ So, Let ϵ<f(y)−m. Then let δ>0 such that x∈I and |x−y|<δ. Then |f(x)−f(y)|<ε<f(y)−m. Thus |f(x)−f(y)|<f(y)−m. Then m-f(y)<f(x)-f(y)<f(y)-m. Then add f(y). Thus m<f(x)<2f(y)-m. Thus there exists δ>0 such that m<f(x) for all x∈I with |x−y|<δ. Is that how it works? $\endgroup$ – Maddy Oct 29 '13 at 3:34
  • $\begingroup$ Where do you get the $ϵ<f(y)−m$? $\endgroup$ – Maddy Oct 31 '13 at 3:18

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