Closure of the algebra and Closure of the set The closure of the set $E$ is defined as $\overline{E} = E \cup E'$, where $E$ is the set of the limit point of $E$.
The closure $\mathscr{B}$ of the algebra $\mathscr{A}$ is defined as the set of the all functions which is the limits of uniformly convergent sequences of members of $\mathscr{A}$.
It seems that $\mathscr{B}$ is only similar to $E'$. How to unify them as consistent?
 A: Admittedly, $\mathscr B$ looks a lot like $E'$. But we can get an "$E$ equivalent" for $\mathscr B$ in this sense: let $a\in\mathscr A$, then the sequence $(a,a,a,\cdots)$ is uniformly convergent, so every $a\in\mathscr A$ can be identified with an element in $\mathscr B$.
My question would be the opposite: why is the $E$ necessary for topological spaces? It really seems like $E'$ should be enough. In fact, it's close to enough; you just need $E$ for the isolated points.
A: Remarks:


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*(Eric Stucky has already given a concise general answer that overlaps with this part.) In general, if $E$ is a subset of a metric space, then $\overline E$ is equal to the set of all limits of sequences in $E$.  This does not mean that all elements of $\overline E$ are limit points of $E$, because limits of sequences from $E$ need not be limit points of $E$.  Example: If $E=\{1,1/2,1/3,\ldots\}$ in $\mathbb R$ with the usual metric, then $\overline E=\{0,1,1/2,1/3,\ldots\}$, and $0$ is the only limit point of $E$.  However, for each positive integer $n$, $\frac1n$ is the limit of the sequence $(\frac1n,\frac1n,\frac1n,\ldots)$ in $E$.  

*If your metric space is the algebra $\ell^\infty(X)$ of bounded real or complex-valued functions on some set $X$ with distance $d(f,g)=\sup_{x\in X}|f(x)-g(x)|$, and $\mathscr A$ is a subalgebra of $\ell^\infty(X)$, then, unless $\mathscr A$ consists of only the $0$ function, every element of $\mathscr A$ is in fact a limit point of $\mathscr A$, so in this case it turns out to be true that $\overline{\mathscr A}=\mathscr A'$.  The reason is that if $0\neq f\in \mathscr A$, then $f$ is the uniform limit of the sequence $((1+1/n)f)_{n=1}^\infty$, and $0$ is the uniform limit of the sequence $(\frac1n f)_{n=1}^\infty$.  
In summary, the two notions are generally consistent without implying that $\overline E=E'$, but in your particular case (if $\mathscr A\neq\{0\}$), or generally for nonzero subspaces of normed vector spaces, $\overline E=E'$ holds, which is equivalent to saying that nonzero subspaces do not have isolated points.
A: You are just taking closures of sets in different metric spaces (using different metrics to define "closure" and "limit point"). You just need to identify in which sets $E$ and $\cal{A}$ lie in, and what metric your associating with each of those two sets.
