Question about Def 7.28 and Thm 7.29 in Baby rudin At first the part in Def 7.28:
"Let $B$ be the set of all functions which are limits of uniformly convergent sequences of members of $A$. Then $B$ is called the uniform closure of $A$.($A$ is family of complex functions on set $E$ and an algebra satisfying $(i)f+g\in A,(ii)fg\in A, (iii)cg\in A,\ \  \forall f,g\in A,c\in \mathbb{C} $)"
and he use this definition to the proof of Thm 7.29 as if this definition have same meaning with the Def 2.26
The part of Thm 7.29:
"By Thm 2.27,  $B$ is (uniformly) closed."
Def 2.26:
If $X$ is a metric space, if $E\subset X$, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of E is the set $\bar{E}=E\cup E'$
Thm 2.27 :
If X is a metric space and $E\subset X$, then $\bar{E}$ is closed.
Why doesn't $B$ have same meaning with $A'$, but have same meaning with $\bar{A}$? 
 A: $B$ is "The set of all functions that are limits of uniformly convergent sequences of members of $A$". We are working in a metric space $C([0,1])$ or $C(X)$ for $X$ compact e.g.) so $B = \overline{A}$: Constant sequences show that $A \subseteq B$ and limits of sequences from $A$ that are not in $A$ already are clearly in $A'$. So $A \cup B \subseteq A \cup A'$, also $A' \subseteq B$ because a limit point of $A$ can always be gotten as the limit of a sequence from $A$.
And the closure of a set $A$ is itself closed (the smallest closed set around $A$).
A: Please point out if there are any errors in the below reasoning regarding uniform closure from Definition 7.28 in Rudin Principles of Mathematical Analysis. Below, we will try to show that closure (Definition 2.26) and uniform closure (Definition 7.28) are identical.
Definition 2.26: If $E\subset X$ of a metric space $X$ and if $E'$ denotes the set of all limit points of $E$ in $X$, then $\bar{E}=E'\cup E$ is the closure of $E$.
Relevant part of Definition 7.28: $\mathscr{B}$ is the set of all functions which are limits of uniformly convergent sequences of members of an algebra $\mathscr{A}$. Then $\mathscr{B}$ is called the uniform closure of $\mathscr{A}$.
First we will show that $\mathscr{A}$ is a subset of $\mathscr{B}$. For any element, $f\in\mathscr{A}$, then we can form a sequence $\left\{ f_{n}\right\}$  taking all $f_{n}$ to be $f$, and this sequence will approach $f$ uniformly, so it will belong to the uniform closure, $\mathscr{B}$. Hence, $\mathscr{A}\subseteq\mathscr{B}$.
Second, we will show that $\mathscr{A}'\subseteq\mathscr{B}$. Any limit point $f$ of $\mathscr{A}$ will have a sequence $\left\{ f_{n}\right\}$  in $\mathscr{A}$ such that $f=\underset{n\rightarrow\infty}{\lim}f_{n}$. [Theorem 3.2 (d)]. Hence, $\mathscr{A}'\subseteq\mathscr{B}$.
The above imply that $\mathscr{A}\cup\mathscr{A}'\subseteq\mathscr{B}\Rightarrow\bar{\mathscr{A}}\subseteq\mathscr{B}$
Lastly, we will show that $\mathscr{B}$ is a subset of the closure of $\mathscr{A}$, that is $\mathscr{B}\subseteq\bar{\mathscr{A}}$. Suppose, $f\in\mathscr{B}$ and $f=\underset{n\rightarrow\infty}{\lim}f_{n}$ where every $f_{n}\in\mathscr{A}$ then we need to show that $f\in\bar{\mathscr{A}}$. If $f\in\mathscr{A}$ then $f\in\bar{\mathscr{A}}$. If $f\notin\mathscr{A}$ then since $f=\underset{n\rightarrow\infty}{\lim}f_{n}$ we have $\forall\varepsilon>0,\exists N\;s.t.\quad n\geq N\Longrightarrow d\left(f_{n},f\right)<\varepsilon$. Here, $d\left(\right)$ is the distance corresponding to the the metric space, which could be the uniform norm or supremum norm (Definition 7.14) but could be some other metric as well. Choose $N$ such that $\quad\varepsilon N>1\Rightarrow1/N<\varepsilon\Rightarrow1/n<\varepsilon$ because $1/n\leq1/N$. Also since, $1/n>1/\left(n+1\right)>1/\left(n+2\right)\ldots$ there are an infinite number of points (functions, $f_{n}$, here) within the neighborhood of radius $\varepsilon$ around $f$. Since this was an arbitrary neighborhood (since $\varepsilon$ was arbitrary), $f$ is a limit point of $\mathscr{A}$ [Theorem 2.20]. Hence, $f\in\mathscr{A}'\subseteq\bar{\mathscr{A}}\Rightarrow\mathscr{B}\subseteq\bar{\mathscr{A}}$.
Hence, we have shown that the closure and uniform closure of $\mathscr{A}$ are the same, that is $\bar{\mathscr{A}}=\mathscr{B}$.
An example where $\mathscr{A}$ is not necessarily closed is the set of polynomials. Consider the algebra of all polynomial functions on an interval $\left[a,b\right]$. Theorem 7.26 (Weierstrass theorem) shows that its uniform closure (or closure) is the algebra of all continuous functions on $\left[a,b\right]$, but the algebra of all polynimials itself does not contain all continuous functions, so it is not closed.
Hence, I am unsure about whether Definition 7.28 requires set $\mathscr{A}$ to be closed or whether $\mathscr{A}$ has to be a subset of a compact metric space (in which case it is closed and bounded). Above, we have tried to show this without explicitly assuming that we are working in a compact metric space or that set $\mathscr{A}$ is closed. Assuming that $\mathscr{A}$ is compact (or closed) might (will?) make it easier, but it does not seem to be necessary.
