I wonder why $\omega$ is continuous iff each $\omega_{i_1,\dots,i_k}$ is continuous. (Calculus on Manifolds by Michael Spivak. continuity of $k$-form)

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I am reading "Calculus on Manifolds" by Michael Spivak.

Many similar considerations may be applied to a function $\omega$ with $\omega(p)\in\Lambda^k(\mathbb{R}_p^n)$; such a function is called a $k$-form on $\mathbb{R}^n$, or simply a differential form. If $\varphi_1(p),\dots,\varphi_n(p)$ is the dual basis to $(e_1)_p,\dots,(e_n)_p$, then $$\omega(p)=\sum_{i_1<\cdots <i_k}\omega_{i_1,\dots,i_k}(p)\cdot [\varphi_{i_1}(p)\wedge\cdots\wedge\varphi_{i_k}(p)]$$ for certain functions $\omega_{i_1,\dots,i_k}$; the form $\omega$ is called continuous, differentiable, etc., if these functions are.

$\omega$ is a function whose value is also a function.
$\omega_{i_1,\dots,i_k}$ is a real-valued function.

I wonder why $\omega$ is continuous iff each $\omega_{i_1,\dots,i_k}$ is continuous.
I wonder why $\omega$ is differentiable iff each $\omega_{i_1,\dots,i_k}$ is differentiable.


I wonder why the author defined $\omega$ is continuous iff each $\omega_{i_1,\dots,i_k}$ is continuous.
I wonder why the author defined $\omega$ is differentiable iff each $\omega_{i_1,\dots,i_k}$ is differentiable.


A function $f:\mathbb{R}^n\to\mathbb{R}$ is called continuous iff each $f_i$ is continuous.
We can prove the following fact:
If $f$ is continuous, then for any positive real number $\varepsilon$, there exists a positive real number $\delta$ such that $|f(x)-f(a)|<\varepsilon$ for any $x$ such that $|x-a|<\delta$.
So I think this definition is natural.

But the author didn't define the distance between $\omega(p_1)$ and $\omega(p_2)$ for arbitrary $p_1,p_2\in\mathbb{R}^n$.