I have some questions on logical implications of the definition of continuity. Here is the context:
- Let $(\Bbb R,d)$ a metric space with the standard metric.
- Define $\,\,f:\Bbb R \rightarrow\Bbb R\,$ to be a function, and let $x_0\in \Bbb R$
- $f$ is said to be continuous at $x_0$ $\iff [\forall\epsilon>0,\,\exists\delta>0;\,\forall x\in \Bbb R,\,|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon].$
My question is: Are the following implications true? I feel like I am mixing up some stuff.
1) $f$ is continuous at $x_0$ $\iff [\forall\epsilon>0,\,\exists\delta>0;\,\forall x\in \Bbb R,\,|x-x_0|<\delta\iff |f(\delta)|\le\epsilon].$ (So that it is sufficient to construct a $\delta$ satsifying $|f(\delta)|\le\epsilon).$
2) $f$ is continuous at $x_0$ $\iff|f(x-x_0)|\le f(x)-f(x_0)$ (I have a feeling I am mis-using the topological definition of continuity.
PARTIAL END: NO NEED TO READ BEYOND (but if you are feeling inspired, then please continue :)
$\\$
If 1) and 2) are true, then to prove continuity, can we do something along the lines of the following:? (Please forgive the complete lack of rigor, I am just trying to run things on a probably false sense of intuition)
Fix $\epsilon=\epsilon_0>0$
We wish to find a $\delta>0$ such that $\forall x\in \Bbb R,\,|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon_0.$
But, there exists $g(x)\in \Bbb R:g(x)*|x-x_0|=|f(x)-f(x_0)|$
But $g(x)*|x-x_0|=|f(x)-f(x_0)|<g(x)*\delta$
And $g(x)*|x-x_0|=|f(x)-f(x_0)|<\epsilon_0$
So we wish to find a $\delta>0$ such that $\delta\le \frac {\epsilon} {g(x)}$
If my last part gives no insight however wrong it may be, please tell me.
You are mixing things up.
The statement
$$ \forall\epsilon>0,\,\exists\delta>0;\,\forall x\in \Bbb R,\,|x-x_0|<\delta\iff |f(\delta)|\le\epsilon.$$ is not equivalent to continuity.
For example the constant function $f(x)=5$ is continuous and $f(\delta)=5$ for every $\delta $
Thus if your $ \epsilon$ is less than $5$ you can not make $$|f(\delta)|\le\epsilon.$$.