Definition of 'dense' in topology Assuming that X is a metric space where $E \in X$, Nr(p) stands for the ball of center p and radius r and E′ is the set of limit points (i.e. the cluster points) of E,
The rudin book I have says 'E is dense in X if every point of X is a limit point of E, or a point of E (or both), so basically if $X = E\cup E'$.  
However, the professor gives the definition as '$E \subset X$ is dense in X if $\forall p\in X, \forall r > 0, \exists q \in E: q \in N_{r}(p) \iff X = E \cup E'$.  
But isn't  $\forall p\in X, \forall r > 0, \exists q \in E: q \in N_{r}(p) \iff X = E'$ ?  
So I thought the professor's definition is not enough for that of 'dense'. Is my understanding correct? If not, I'd like to know where the flaw is. Thank you in advance.
 A: Normally the definition of a limit point requires that $p\neq q$ in your nomenclature. For example, if $X=[1,2]$, and $E=\{1\}$, then $E'=\emptyset$, $1$ isn't a limit point of $E$, but of course every neighborhood of $1$ contains $1$.
So for a more complicated example, if $X=\{0\} \cup [1,2]$ and $E=\{0\} \cup ([1,2] \cap \mathbb Q)$. You want $E$ to be dense in $X$, but $0$ is not going to be a limit point of any subset of $X$. That's why you "have to" include $E$ as well, to get the "usual sense" of dense.
A: If by $E'$ you mean the set of limit points of $E$, then the statement
$$\forall p\in X, \forall r > 0, \exists q \in E: q \in N_{r}(p) \iff X = E'$$
is wrong. For example, take $X=\mathbb R$ with the standard metric, and $E=\{0\}$.
Then, $E'=\emptyset$, but the set $X=\{0\}$ satisfies the condition
$$\forall p\in X, \forall r > 0, \exists q \in E: q \in N_{r}(p)$$
This is because, if $p\in X$, then $x=0$. Then, if $r>0$, we can take $q=0$, and have $q\in N_r(p)$.

The definition of a limit point is:

For a set $E$, $p$ is a limit point of $E$ if, for every $r>0$, there exists some $q\in E$ such that $q\in N_r(p)$ and $q\neq p$.

That last condition, $q\neq p$, is important.
A: So after all, it seems like the professor was sort of right in what he said and my equivalence was flawed.
First of all,
$\forall p \in X, \forall r>0, \exists q \in E: q \in N_{r}(p) \wedge q \neq p \iff X \subseteq E'$ , as pointed out by multiple answers.  So the condition $p \neq q$ is obviously crucial in the above equivalence.  
Next, regarding the professor's definition, he was right in saying that 
$\forall p \in X, \forall r>0, \exists q \in E: q \in N_r(p) \iff$ E is dense in X.   It is because $\forall p \in X, \forall r>0, \exists q \in E: q \in N_{r}(p) \wedge q \neq p \iff X \subseteq E' \cdots (a)$ 
$\forall p \in X, \forall r>0, \exists q \in E: q \in N_{r}(p) \wedge q=p \iff \forall p \in X, p \in E \iff X \subseteq E \cdots (b)$ 
With (a) and (b) combined, then  $\forall p \in X, \forall r>0, \exists q \in E: q \in N_{r}(p) \iff X \subseteq E \cup E'$,  which means that every element of X is either a limit point of E or an element of E (or both), which fits in the Rudin's definition of density.
