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mathmari
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Suppose that $h_1,h_2:[0,1]\rightarrow \mathbb{R}$ be continuous functions.
I want to show that there is a $\epsilon>0$ such that $|h_1(x)-h_2(x)|\geq \epsilon$ for each $x\in [0,1]$, given that $h_1(x)\neq h_2(x)$ for each $x\in [0,1]$. I have done the following:
We know that $h_1(x)\neq h_2(x)$ for each $x\in [0,1]$, that means that $f(x):=h_1(x)-h_2(x)$ is either $>0$ or $<0$ but never $=0$.
Therefore, we have that $|f(x)|>0, \forall x\in [0,1]$. That means that there is a $\epsilon>0$ such that $|f(x)|\geq \epsilon, \forall x\in [0,1]$. Is this correct? (Wondering)
Suppose that $h_1,h_2:[0,1]\rightarrow \mathbb{R}$ be continuous functions.
I want to show that there is a $\epsilon>0$ such that $|h_1(x)-h_2(x)|\geq \epsilon$ for each $x\in [0,1]$, given that $h_1(x)\neq h_2(x)$ for each $x\in [0,1]$. I have done the following:
We know that $h_1(x)\neq h_2(x)$ for each $x\in [0,1]$, that means that $f(x):=h_1(x)-h_2(x)$ is either $>0$ or $<0$ but never $=0$.
Therefore, we have that $|f(x)|>0, \forall x\in [0,1]$. That means that there is a $\epsilon>0$ such that $|f(x)|\geq \epsilon, \forall x\in [0,1]$. Is this correct? (Wondering)
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