Resolvent: Definition Given a Banach space $E$.
Consider linear operators:
$$T:E\supset\mathcal{D}(T)\to E:\quad T(\kappa x+\lambda y)=\kappa T(x)+\lambda T(y)$$
(No other assumptions on the operator!)
Denote for shorthand:
$$R_\lambda:\mathcal{R}_\lambda\to\mathcal{D}_\lambda:\quad R_\lambda:=(\lambda-T)^{-1}$$
The standard definition for the resolvent set:
$$\rho(T):=\{\lambda\in\mathbb{C}:\mathcal{N}_\lambda=(0),\mathcal{R}_\lambda=E,\|R_\lambda \|<\infty\}$$
An alternative definition for the resolvent set:
$$\rho(T):=\{\lambda\in\mathbb{C}:\mathcal{N}_\lambda=(0),\overline{\mathcal{R}_\lambda}=E,\|R_\lambda\|<\infty\}$$
(See Werner or Weidmann resp. Kubrusly or Kreyszig.)
Do these definitions really agree?
Clearly for closed operators they do.
 A: Here's an idea. Let $X=l^{2}$ be the space of sequences $\{ x_{n}\}_{n=1}^{\infty}$ of square-summable sequences of complex numbers. Let $e_{n}$ the standard basis element which is a sequence with all 0's except for a $1$ in the n-th place. Let $M$ be the subspace spanned by finite linear combinations of $\{ e_{1}+\frac{1}{n}e_{n}\}_{n=2}^{\infty}$. The closure $M^{c}$ includes $e_{1}$ and, because of that, also includes every $e_{n}$ for $n \ge 2$. So $M^{c}=X$. Define $S$ to be the restriction of the left shift to $M$. $S$ is bounded on $M$, and $\mathcal{N}(S)=\{0\}$ because $e_{1} \notin M$. Let $T$ be the inverse of $S$. Then $T : \mathcal{D}(T) \rightarrow X$ has domain $\mathcal{D}(T)=S(M)$, and the range of $T$ is $M$ which is dense in $X$. And $S=T^{-1}$ is bounded by $1$, which gives $\|Tx\| \ge \|x\|$ for all $x \in \mathcal{D}(T)$. So $0 \in \rho(T)$ using the second definition of resolvent, which is very wrong from my point of view. However, $T$ is not closable because $(e_{1},0)$ is in the closure of the graph of $S$ and, so, $(0,e_{1})$ is in the closure of the graph of $T$. I think this works, but please check my details.
A: The business is rather wolly but basically everything boils down to closed operators...

Before, let's set our framework:
Framework:
As always we consider Banach spaces: $X$, $Y$
First let's define the resolvent set properly:
Definition:
The resolvent set is defined as those values where the linear problem can be solved always, uniquely and continuously:
$$\rho(T):=\{\lambda\in\mathbb{C}:\mathcal{N}(T-\lambda)=\{0\},\mathcal{R}(T-\lambda)=Y,C\|(T-\lambda)x\|\geq\|x\|\}$$
Next let's state some general theorems on closed operators:
Theorem 1: 
A continuous operator with closed domain is closed:
$$\mathcal{D}(T)\text{ closed}:\quad T\text{ continuous}\Rightarrow T\text{ closed}$$
Theorem 2:
A closed operator with complete codomain is continuous:
$$\mathcal{D}(T)\text{ complete}:\quad T\text{ closed}\Rightarrow T\text{ continuous}$$
Theorem 3:
The domain of a continuous closed operator with complete codomain is closed:
$$T\text{ closed, continuous}\Rightarrow\mathcal{D}(T)\text{ closed}\qquad(\mathcal{C}(T)\text{ complete})$$
Now let's investigate our situations:
Consequence 1:
The resolvent set of unclosed operators is empty:
$$T\text{ not closed}\Rightarrow\rho(T)=\varnothing$$
Observation:
A cobound automatically guarantees the existence of the resolvent:
$$C\|(T-\lambda)x\|\geq\|x\|\Rightarrow\mathcal{N}(T-\lambda)=\{0\}$$
Consequence 3:
The domain of a continuous resolvent is automatically closed:
$$T\text{ closed}:\quad C\|(T-\lambda)x\|\geq\|x\|\Rightarrow\mathcal{R}(T-\lambda)\text{ closed}$$
Consequence 2:
An everywhere defined resolvent is automatically continuous:
$$T\text{ closed}:\quad\mathcal{N}(T-\lambda)=\{0\},\mathcal{R}(T-\lambda)\text{ closed}\Rightarrow C\|(T-\lambda)x\|\geq\|x\|$$
Finally let's conclude:
Conclusion:
An alternative definition therefore could address the resolvent:
$$R(\lambda)\text{ exists, densely defined, closely defined, continuous}$$
So spectrum then could be splits up into:
$$\lambda\in\sigma_p(T):\iff R(\lambda)\text{ not exists}$$
$$\lambda\in\sigma_r(T):\iff R(\lambda)\text{ exists, not densely defined}$$
$$\lambda\in\sigma_c(T):\iff R(\lambda)\text{ exists, densely defined, not closely defined}$$
and the resolvent set becomes:
$$\lambda\in\rho(T):\iff R(\lambda)\text{ exists, everywhere defined, continuous}$$
Note that the remaining combination for closed operators does not occur:
$$T\text{ closed}: R(\lambda)\text{ exists, everywhere defined}\Rightarrow R(\lambda)\text{ continuous}$$
Outlook:
A more detailed distinction is listed in Kubrusly's 'The Elements of Operator Theory'.
