Prove that $f(x) = \frac{1}{x}$ is not uniformly continuous on $(0,1)$

Question

I want to know if this proof is rigorous enough.

What we need to show here is that ∃ε>0 ∀δ>0 such that |f(x) - f(y)|>ε ,∃x,y∈(0,1) where |x-y|<δ
Let ε=1
    Let δ be given.
        Case 1: δ ≥ 1,
            let x = 1/δ , y = 1/3δ
            then we have |x-y|=|2/3δ|<δ
            further |f(x)-f(y)| = |-2δ|> δ ≥ 1 = ε
        Case 2: δ < 1,
            let x = δ , y = δ/2
            then we have |x - y|=|δ/2| < δ
            further |f(x) - f(y)| = |1/δ - 2/δ| = 1/δ > 1 =  ε

Hence in both cases we have shown that there is some ε for which, no matter how close we pick two points in the domain (0,1) the distance between the images under the function is not within ε of each other.

P.S - There were a few other threads stating this proof, but none were similar to my method plus some requires other theorems I'm yet to learn. Hope this is not a duplicate. Appreciate your time. Thanks in advance.

Answer 1

Your proof is invalid in the specific case of $\delta=1$, as $x=1$ is illegal. You need to handle that separately if you're going to keep the rest of your proof untouched. Alternately, you could instead use $x=\frac1{2\delta}, y=\frac1{4\delta}$ in the first case. Alternately, you could use $x=\frac12, y=\frac14$, not caring about $\delta$ at all.

If you do decide to handle $\delta=1$ separately, you could without issues use $y=\frac1{2\delta}$ in the first case, to make your two cases more symmetric. This is an entirely unnecessary change, but more of an aesthetic concern.