Suppose we have an operator $A \in B(X,Y)$ where $X$ is a subset of $Y$ (X and Y are Hilbert spaces). Does the fact that $\beta I - A$ is bijective implies that $\beta \in \rho(A)$?

A is closed as an operator from $X$ to $Y$ since it's bounded. Also, for a closed operator $\beta I - A$ bijective implies $\beta \in \rho(A)$.

I feel that something is not ok with this reasoning, but I can't tell where.

Suppose we have an operator $A \in B(X,Y)$ where $X$ is a subset of $Y$ (X and Y are Hilbert spaces). Does the fact that $\beta I - A$ is bijective imply that $\beta \in \rho(A)$?

A is closed as an operator from $X$ to $Y$ since it's bounded. Also, for a closed operator $\beta I - A$ bijective implies $\beta \in \rho(A)$.

I feel that something is not ok with this reasoning, but I can't tell where.

Suppose we have an operator $A \in B(X,Y)$ where $X$ is a subset of $Y$. Does the fact that $\beta I - A$ is bijective implies that $\beta \in \rho(A)$?

A is closed as an operator from $X$ to $Y$ since it's bounded. Also, for a closed operator $\beta I - A$ bijective implies $\beta \in \rho(A)$.

I feel that something is not ok with this reasoning, but I can't tell where.

Suppose we have an operator $A \in B(X,Y)$ where $X$ is a subset of $Y$ (X and Y are Hilbert spaces). Does the fact that $\beta I - A$ is bijective implies that $\beta \in \rho(A)$?

A is closed as an operator from $X$ to $Y$ since it's bounded. Also, for a closed operator $\beta I - A$ bijective implies $\beta \in \rho(A)$.

I feel that something is not ok with this reasoning, but I can't tell where.