On the spectrum of a bounded linear operator According to [wikipedia][1]

Let $T$ be a bounded linear operator acting on a Banach space $X$ over the complex scalar field $\mathbb{C}$ and $I$ be the identity operator on $X$. The spectrum of $T$ is the set of all $\lambda \in \mathbb{C}$ for which the operator $T-\lambda I$ does not have an inverse that is a bounded linear operator

This definition seems a like unprecise to me because of the following. Because $X$ is Banach, if $T$ has an inverse, [this inverse must be bounded][2]. But (in my opinion) the definition on wikipedia might be misleading because one could think that it could happen that $T-\lambda I$ is invertible but not bounded, in which case $\lambda$ seems also to be an element of the spectrum of $T$ according to the above definition. I think a better definition of the spectrum, in this case, would be the set of all complex numbers such as $T-\lambda I$ is not invertible.
Question: If $X$ is assumed to be normed instead of Banach, what is the best definition of spectrum? Does one demand $T-\lambda I$ not to be invertible or not to be invertible and bounded?
[1]: https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#:~:text=%2C%20for%20all%20).-,Basic%20properties,subset%20of%20the%20complex%20plane.&text=would%20be%20defined%20everywhere%20on%20the%20complex%20plane%20and%20bounded.&text=The%20boundedness%20of%20the%20spectrum,bounded%20by%20%7C%7CT%7C%7C.
[2]: The inverse of bounded operator?
 A: If $T-\lambda I$ is injective, then $T-\lambda I$ will have an inverse on $\mathcal{R}(T-\lambda I)$, but that does not guarantee that $(T-\lambda I)^{-1} : \mathcal{R}(T-\lambda I)\subset X\rightarrow X$ is bounded. For example, consider $T : L^2[0,1]\rightarrow L^2[0,1]$ defined by
$$
              Tf = \int_0^x f(t)dt.
$$
$T$ is bounded. Even though the inverse $T^{-1}g = g'$ is closed, it is defined only on functions $g \in L^2[0,1]$ that are
$\;\;\;$(i) absolutely continuous,
$\;\;\;$(ii) vanish at $0$, and
$\;\;\;$(iii) have a square-integrable derivative on $[0,1]$.
Furthermore $T^{-1}$ is not bounded on its domain; so it is not possible to extended $T^{-1}$ in such a way that it will be continuous. If the range of $T$ were all of $X$, so that the inverse of $T$ were defined everywhere on $L^2[0,1]$, then your argument would apply because $T$ would be defined on a Banach space and would have a closed graph. But that doesn't have to happen, even if $T^{-1}$ exists, as it does not happen in this case.
A: The issue is an injective bounded operator $T\in\mathcal{B}(X)$ might not be surjective, and therefore its set-theoretic inverse $T^{-1}: T(X)\to X$ might not be defined everywhere on $X$.
You are right in that, if $T$ is bijective and bounded, then, by the open mapping theorem, $T^{-1}$ is well-defined everywhere on $X$ and is also bounded. But the definition of the spectrum leaves open all possible scenarios. It could be $T$ is bounded but not injective, or injective but its image is smaller than the whole space (in which case, there are the possibilities of the image being closed or dense or contained within a proper closed subspace).
