How to prove that $(E\cup d(E))^c$ is open? I am trying to understand my teacher's proof of how $(E\cup d(E))^c$ is open.  This is the proof
Let $$x\in (E\cup d(E))^c$$
Then $$x\not\in (E\cup d(E))$$
which means $x\not\in E $ and $x\not\in  d(E)$. Since $x\not\in  d(E)$ then $\exists G_{x}\in T\ni (G_{x}\cap E)-\{x\}=\phi$. Hence any point of $G_{x}$ not limit point of $E$.
$$G_x\subseteq (d(E))^c$$
The rest of the proof I understand.
My problem is that I don't understand how my teacher got the statement "Hence any point of $G_{x}$ not limit point of $E$" because from what I understand is that if there is $G_{x}\in T$ such that $(G_{x}\cap E)-\{x\}=\phi$ then $x$ not a limit point of $E$ but in the proof my teacher said every point in $G_{x}$ (not just $x$) is not limit $E$. Does anyone know why this is?
 A: Let $$x\in (E\cup d(E))^c$$
Then $$x\not\in (E\cup d(E))$$
which means $x\not\in E $ and $x\not\in  d(E)$. Since $x\not\in  d(E)$ then $\exists G_{x}\in T$ such that $$\Big(G_{x}\smallsetminus\{x\}\Big)\cap E=\phi.$$
Since $x\notin E$, we conclude that $$G_x\cap E=\phi.\tag 1$$ Next, we will show that $G_x\cap d(E)=\phi.$ Suppose that $G_x\cap d(E)\neq\phi.$ Let $g\in G_x\cap d(E)$. Since $G_x$ is open and $g\in G_x$, and $g\in d(E)$, we conclude that
$$\Big(G_x\smallsetminus \{g\}\Big)\cap E\neq \phi.$$ But
$$\phi\subset\Bigg[\Big(G_x\smallsetminus \{g\}\Big)\cap E\Bigg]\subset G_x\cap E=\phi,$$ implying that 
$$\Big(G_x\smallsetminus \{g\}\Big)\cap E=\phi.$$ We obtain a contradiction. Hence, $$G_x\cap d(E)=\phi.\tag 2$$ Now, $(1)$ and $(2)$ both implies that 
$$G_x\subset E^c$$ and $$G_x\subset [d(E)]^c.$$ Hence,$$G_x\subset E^c\cap [d(E)]^c$$ which means that $$x\in G_x\subset(E\cup d(E))^c.$$ Done.--)
A: Consider any point in $y\in G_x$. The definition of convergence in a topological space is as follows:
Definition: A sequence $(y_t)_{t=1}^{\infty}\subseteq X$ converges to $y$ if for every open set $U$ containing $y$, there exists some $N_U\in \mathbb{N}$ such that $t>N_U$ implies $y_t\in U$. 
In other words, a sequence $(y_t)_{t=1}^{\infty}$ converges to a point $y$ if for every open neighborhood U of $y$, the sequence $(y_t)_{t=1}^{\infty}$ is eventually contained inside of U.
So back to your example. Take some $y\in G_x$. Assume to the contrary that there was some sequence of points $(x_t)_{t=1}^\infty\subseteq E$ that converged to $y$. From the definition of convergence, every open neighborhood of $y$ must eventually contain points in the sequence. However, from what you have shown, $E\cap G_x = \emptyset$. Hence, you have identified an open set of $y$ (namely $G_x$) such that $(x_t)_{t=1}^\infty$ is never contained inside of it. This is a contradiction. Hence $y$ (and thus any point in $G_x$) is not a limit point of $E$. 
