Properties of the $C^{*}$-algebra $\mathbb{C}\oplus \mathcal{U}$ Let $\mathcal{U}$ be a $C^{*}$-algebra without a unit and consider the algebra $\mathbb{C}\oplus \mathcal{U}$ formed by the ordered pairs $(\alpha, A)$, $\alpha \in \mathbb{C}$ and $A \in \mathcal{U}$ with the operations $(\alpha, A) + (\beta, B) := (\alpha+\beta, A+B)$ and $(\alpha,A)(\beta, B) := (\alpha \beta, \alpha B + \beta A + AB)$. With these notations, $\mathcal{U}$ is identifiable with the subalgebra formed by pairs $(0, A)$, $A \in \mathcal{U}$.
Bratteli and Robinsons's book states the following facts:
(1) $\mathbb{C}\oplus \mathcal{U}$ is a $C^{*}$-algebra with unit ${\bf{1}}$ and,
(2) If $\mathcal{U}$ is a $C^{*}$-algebra with unit ${\bf{1}}$, it is possible to have a $C^{*}$-subalgebra $\mathcal{B}$ of $\mathcal{U}$ without unit. In this case, the $C^{*}$-algebra $\mathbb{C}\oplus \mathcal{B}$ obtained by adjoining the unit to $\mathcal{B}$ is the smallest $C^{*}$-subalgebra which contains both $\mathcal{B}$ and ${\bf{1}}$.
Now, about these statements:
Question 1: If $\mathcal{U}$ does not have a unit, I suppose the larger algebra $\mathbb{C}\oplus \mathcal{U}$ has the unit ${\bf{1}} = (1,0)$, right? However, if $\mathcal{U}$ has a unit and $\mathcal{B}$ is a subalgebra that does not, the unit on $\mathbb{C}\oplus \mathcal{U}$ would be $(1,0)$ and not ${\bf{1}}$? How do these two units relate?
Question 2: What exactly does statemente (2) mean? Does "smallest $C^{*}$-algebra" mean that any $C^{*}$-algebra satisfying the same properties must contain $\mathbb{C}\oplus \mathcal{U}$? Also, how does one garantee that this is, in fact, the smallest?
 A: For your question 2:
Let $\mathcal{U}$ be a $\text{C}^{*}$-algebra with a unit $\mathbf{1}$, and let $\mathcal{B} \subset \mathcal{U}$ be a $\text{C}^{*}$-subalgebra, which does not contain the unit $\mathbf{1} \in \mathcal{U}$.
First, we must understand in what sense $\mathbb{C} \oplus \mathcal{B}$ is a subalgebra of $\mathcal{U}$. This is contained in the following claim:

The map $\mathbb{C} \oplus \mathcal{B} \rightarrow \mathcal{U}, (\lambda, b) \mapsto \lambda\mathbf{1} + b$ is an injective $\text{C}^{*}$-homomorphism.

Now, it is clear that $\mathbb{C} \oplus \mathcal{B}$ contains both $\textbf{1} \in \mathcal{U}$ and $\mathcal{B}$.
The claim that $\mathbb{C} \oplus \mathcal{B}$ is the smallest such algebra, means that if $\mathcal{A} \subset \mathcal{U}$ is a $\text{C}^{*}$-subalgebra which contains both $\mathbf{1}$ and $\mathcal{B}$, then $\mathbb{C} \oplus \mathcal{B} \subseteq \mathcal{A}$.
For question 1:
The units do not relate.
Suppose that $\mathcal{U}$ is a $\text{C}^{*}$-algebra with unit, and that $\mathcal{B} \subset \mathcal{U}$ is a $\text{C}^{*}$-subalgebra which does contain the unit $\mathbf{1} \in \mathcal{U}$.
In this case, the map $(\lambda, b) \mapsto \lambda \mathbf{1} + b$ is not injective (exercise: find the kernel); and thus $\mathbb{C} \oplus \mathcal{B}$ is not a subalgebra of $\mathcal{U}$.
Now, we should observe that even though $\mathcal{U}$ is a subalgebra of $\mathbb{C} \oplus \mathcal{U}$ which contains a unit $\mathbf{1} \in \mathcal{U}$, it certainly does not contain the unit $(1,0) \in \mathbb{C} \oplus \mathcal{U}$.
