Continuity of norm. Need to understand how and why

Question

$f:X \to \mathbb R \ \ \ , \ f(x)=\| x\|.$ Prove that $f$ is continuous.

I have this definition of continuity in metric spaces:

1. 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:

2. $\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 $

3. $f \in C(a) \text{ iff } f^{-1}(V) \in \tau \text{ if } V \in \tau \text{ or if every inverse image of a open set is open.}$

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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.

Answer 1

Concerning the usage of "if":

In definitions people usually write X is called Y if Z holds, even though they mean "if and only if". Since it is a definition, there is no other object with that name and other probably weaker definitions.

Concerning "if", "iff", and $\implies$:

"If condition A holds, then statement B is valid" is usually written as $A \implies B$ and read as $A$ implies $B$. "Iff", short for "if and only if", means $A \implies B$ and $B\implies A$ are true at the same time.

Concerning the characterizations of continuity:

Here, $f^{-1}$ means the set of all inverse images: $$ f^{-1}(V) = \{ x\in X \mid f(x) \in V \}. $$ It is always defined, but might be empty.

Definitions 1 and 2 are the same in metric space. To see this, each neighborhood of $a$ contains some $\delta$-neighborhood of $a$ as subset.

Definition 3 states that $f$ is continuous at any $a\in X$, not only at a particular one, like in definition 2. To see that "continuous at every $a\in X$" as in definition 2 is the same as definition 3, a set $S\subseteq X$ is open if and only if each $x\in S$ is an interior point of $S$.

Concerning your proof:

For definition 1, you're are absolutely right. In a formal proof however, for any given $\varepsilon > 0$ you have to pick a $\delta > 0$ (may depending on $\varepsilon$).

For definition 2, you have a typo there. $\{ \| x \| \mid x\in B(a,\delta) \}$ is a subset of $(\|a\| - \varepsilon, \|a\| + \varepsilon)$.