Limit proof by contrapositive

111 Views Asked by At

I'm writing a proof for the statement : "If a function $f$ has a limit at $p$, then for every $\epsilon >0$ there is a $\delta>0$ such that $\vert f(x_1) - f(x_2)\vert < \epsilon$, whenever $x_1,x_2 \in U'(p,\delta)$."

I'm doing this by contrapositive and I wanted to make sure I have written the structure correctly. This means that for $\delta >0$ there is a $\epsilon >0$ such that $$\vert f(x_1)-f(x_2) \vert \geq \epsilon$$

whenever $x_1,x_2 \in [U'(p,\delta)]^c$. The proof then requires that we show there exists a $\epsilon>0$ such that for all $\delta >0$ there is a point $x$ ( which depends on $\delta$ ), for which $$\vert f(x) - L \vert \geq \epsilon$$

but $$\vert x-p \vert < \delta.$$

$U'(p,\delta)$ is a punctured neighborhood of $p$. Is this logically sound?

1

There are 1 best solutions below

2
On BEST ANSWER

The correct negation is:

There exists $\epsilon >0$ such that for every $\delta>0, $ there exists $x_1,x_2 \in U'(p,\delta)$ satisfying $$|f(x_1)-f(x_2)|\geq \epsilon.$$