Prove that if $f(x)$ has a greatest lower bound $L$, then for every $\epsilon\gt 0$ there exists an $\bar{x}$ in the domain of $f(x)$ such that $|f(\bar{x})-L|\lt\epsilon$.

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Lesson Parent: 
Geometry of Functions II: The Extreme-Value Theorem
Problem: 

Prove that if $f(x)$ has a greatest lower bound $L$, then for every $\epsilon\gt 0$ there exists an $\bar{x}$ in the domain of $f(x)$ such that $|f(\bar{x})-L|\lt\epsilon$.

Answer: 

It is true that if $f(x)$ has a greatest lower bound $L$, then for every $\epsilon\gt 0$ there exists an $\bar{x}$ in the domain of $f(x)$ such that $|f(\bar{x})-L|\lt\epsilon$.

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Solution: 

Since $L$ is an lower bound then for every $x$ in the domain of $f(x)$ then <Sign in to see all the formulas>.

Let's denote $n$ as any positive integer. Now since $L$ is also the greatest lower bound then $L+1/n$ cannot be a lower bound. Since $L+1/n$ is not a lower bound, there exists <Sign in to see all the formulas> in the domain of $f(x)$ such that <Sign in to see all the formulas>.

Therefore, for every positive integer $n$ there exists an <Sign in to see all the formulas> such that <Sign in to see all the formulas>. By subtracting $L$ we see that <Sign in to see all the formulas>. Therefore, <Sign in to see all the formulas>.

Now we know that for every <Sign in to see all the formulas> there exists a natural number $n$ such that <Sign in to see all the formulas>. So now we know that for every <Sign in to see all the formulas> there exists an <Sign in to see all the formulas> such that <Sign in to see all the formulas>.