$f:X \to \mathbb R \ \ \ , \ f(x)=\| x\|.$ Prove that $f$ is continuous.
I have this definition of continuity in metric spaces:
- Let $(X, d_x)$ and $(Y,d_y)$ be metric spaces. $$f\in C(a) \text{ if } (\forall \mathcal{E} > 0)(\exists \delta > 0)(d_x(a,y)< \delta ) \implies d_y(f(a),f(y))< \mathcal{E}$$
And the following definition for continuity in topological spaces:
$\text{For }(X, \tau ) \text{, } (Y, \tau') \text{, let} f: X \to Y \text{ and } a \in X \text{. Then } f \in C(a) \\ \text{a) If } \forall V_{f(a)}: \ \exists U_a \text{ (these are neighboorhoods of f(a) and a respectively) such that }\\ f(U_a)\subseteq V_{f(a)} \\ \text{b) If } \forall (V_{f(a)}): \ \exists U_a : f^{-1}({V_{f(a)}})=U_a $
$f \in C(a) \text{ iff } f^{-1}(V) \in \tau \text{ if } V \in \tau \text{ or if every inverse image of a open set is open.}$
Questions:
For 1.:
If the $(\forall \mathcal{E})$ (the condition in 1.) is true, does that mean that the function is continuous in the metric space? This is confusing me because it is just an "if " condition. If this is true, I tried proving continuity of the norm like this: $\| \cdot\|$ generates $d_x = \|x-y \|.$ So I have $|f(x)-f(y)|=|\|x \|-\| y\||\leq \|x-y \|$, and it would be easy to see how to apply the first definition.
For 2.:
I'm guessing they used this to prove the hypothesis:$|f(x)-f(y)|=|\|x \|-\| y\||\leq \|x-y \|$ . Then (at this point I don't understand which definition is used) this implies that $\forall a \in X$, $\forall \varepsilon >0, 0< \delta < \varepsilon \ B(a, \delta)$ lays in the $\varepsilon-$neighboorhood $(\| a\|- \varepsilon, \|a \|+ \varepsilon)$ of point $a$.
For 2. and 3.:
When do I use these theorems with $f^{-1}$? Not all functions have an inverse.
The thing that causes me the most confusion is when "if" and "iff" are used , I am not convinced whether "if" means "$\implies$" because what I wrote seems invalid. If I have a wrong understanding of these logical operations, I'd like someone to enlighten me in that regard too.