How to negate: not a limit point (symbolic logic) 1.There are few I have seen here.
$\forall N(x), \exists x'\in B, (x'\neq x\wedge x'\in E)$.
$\forall N(x), \exists x'\in B, (x'\neq x\to x\not\in E)$.
$\forall r>0, \exists x'\in N_r(x)\cap E, x'\neq x$
$\forall r>0, d(x',x)<r$
...etc? 
(can't make it a full sentence for the last one using distance)
How should I negate these statements?
2.What about an interior point? (how to negate)
$\exists r>0, x'\in N_r(x)\to x'\in E$
3.Finally, how should I use these to prove that $E$ is open iff $E^c$ is closed? (Rudin 2.33)
Suppose $E^c$ is closed
Proof:
Choose $x\in{E}$, so $x\not\in{E^c}$
{By the contrapositive of the definition of a closed set:
$x\in{E^c}'\to x\in E^c$ (by definition)
$x\not\in{E^c}\to x\not\in{E^c}'$ (by contrapositive)
}
$x\not\in{E^c}'$
$x\in\{??\}$ negation of limit point
....
 A: A point $x$ is a limit point of $E$, whenever for every neighbourhood $O$ of $x$, there is some $x' \in E$, such that $x \neq x'$ and $x' \in O$.
So in a first order formula, assuming we have the notation $\mathcal{N}(x)$ for the set of all (or just the open) neighbourhoods of $x$:
$$x \in E' \text{ iff } \forall O \in \mathcal{N}(x): \exists x' \in E: (\lnot(x = x') \land x' \in O)$$
The negation of this  using standard logical rules would be 
$$ x \notin E' \text{ iff } \exists O \in \mathcal{N}(x): \forall x' \in E: x = x' \lor x' \notin O$$
And the last part can be seen to be equivalent to $E \setminus \{x\} \subseteq X \setminus O$ or $O \cap (E \setminus \{x\}) = \emptyset$.
So $$ x \notin E' \text{ iff } \exists O \in \mathcal{N}(x): O \cap (E \setminus \{x\}) = \emptyset$$
A point $x$ is an interior point of $E$ iff $\exists O \in \mathcal{N}(x): O \subseteq E$, so it's not iff $\forall O \in \mathcal{N}(x): O \nsubseteq E$, and the latter means that this $O$ intersects $X \setminus E$ (some point in $O$ is not in $E$, which is exactly such an intersection point).
So $$x \notin \operatorname{Int}(E) \text{ iff } \forall O \in \mathcal{N}(x): O \cap (X \setminus E) \neq \emptyset $$
To see that $E$ is open iff $X \setminus E$ is closed, these formulae are not too much of a help. Just reasoning will do it: suppose $E$ is open. To see that $F = X \setminus E$ is closed, we need to prove $F' \subseteq F$. So suppose $x \in F'$, and $x \notin F$, so $x \in E$. As $E$ is open and contains $x$ it is a neighbourhood of $x$ that howver misses $F$ entirely while $x \notin F$. This would mean that $x \notin F'$ (you could use the formulae above to see this), contradiction. So such an $x$ cannot exist and $X \setminus E$ is closed. 
The other direction is quite similar as well.
A: To negate the given statements, first apply the rule of dual negation of the quantifiers, repeating as needed until the negation sign is "moved" to the predicate, and then negate the predicate, using DeMorgan's rules if required.
The negation of "every $x$ in $A$ is $P(x)$" is "some $x$ in $A$ is not $P(x)$"
$$\neg\Big(\forall x \in A, P(x)\Big) \; \iff \; \exists x\in A, \neg P(x)$$
And vice versa:
$$\neg\Big(\exists x \in A, P(x)\Big) \; \iff \; \forall x\in A, \neg P(x)$$
Note: We do not negate the restriction on the domain of discourse.
So, for example:
$$\begin{align}
\neg \forall x \in A, \exists y\in B, \big(y=f(x) \vee x=f(y)\big)
\; \iff \; & \exists x\in A, \neg \exists y\in B, \big(y=f(x) \vee x=f(y)\big) 
\\ \iff \; & \exists x\in A,\forall y\in B, \neg \big(y=f(x) \vee x=f(y)\big)
\\ \iff \; & \exists x\in A,\forall y\in B, \big(y\neq f(x) \wedge x\neq f(x)\big)
\end{align}$$

Also $\neg (A\to B) \;\iff\; (A\wedge \neg B)$
Can you do the first and second problems now?

For the third problem, hint: a set is closed if it contains all of its limits points, and a set is open if it contains none of them.  So show that all the limit points of a set are limit points of the complement and vice versa.
