Confusion with Closures in the topological sense The rigorous definition is

A closure of $A \subseteq X$ of a topological space $X$ is denoted Cl($A$) and is the intersection of all closed subsets of $X$ that contain $A$.

The more intuitive or loose but correct explanation and my understanding is,

Closure of $A$ is essentially the smallest possible closed subset of $X$ that contains $A$.

Now, there seems to be many ways to rephrase or think about this notion of closure. One such I cannot hold my doubts against is,

Cl($A$)$=\{x \in X:$every neighborhood of $x$ meets $A$ $\}$. $*$Given $A,B \subseteq X$ we say $A$ meets $B$ if $A \cap B \neq \phi$.

Let's look at a specific example that buttresses my doubts on this one.
I have a set $X=\{a,b,c\}$ so three point set. I define a topology on this as follows $\tau=\{X,\phi, \{b,c\},\{a\}\}$. Let me pick the subset $A=\{c\} \subseteq X$. What is the closure of this $A$?
Well, I only have three elements to choose so let me do it step by step.
$a$... what is its neighborhood? A neighborhood of $a$ is a subset of $X$ that contains an open neighborhood of $a$. Looking at $\tau$, I see that $\{a\}$ and $X$ are open neighborhoods of $a$. So this must be in the neighborhood of $a$. However, intersection $\{a\} \cap A = \phi$ so this immediately tells me $a \not\in$ Cl$(A)$.
Let's move on to $b$. Then open neighborhoods are $X,\{b,c\}$. Thus the intersection with the neighborhoods of $b$ must mean I intersect either of these two with $A$ and in both cases, $X \cap A =\{c\} \neq \phi$ and $\{b,c\} \cap A =\{c\} \neq \phi$. Thus $b \in$ Cl$(A)$.
Finally, $c$. The open neighborhoods are $X,\{b,c\}$ just as for $\{b\}$. Thus the intersection with its neighborhood yields exactly the same result as in for the case in $b$ above. i.e. $c \in$ Cl$(A)$ (well, it's obvious this is true since the closure must contain $A=\{c\}$).
But then, putting the result together, I conclude that $b,c \in$Cl$(A)$. But then, Cl$(A)=\{b,c\} \in \tau$ thus Cl$(A)$ is open. However, it needs to be closed, so a contradiction by definition.
Big question mark; what did I do wrong? Whenever I do topology, I always stumble across these mind boggling contradictions and (most likely wrong but seemingly true) counter examples without fail that keep me irritated for hours because I can't spot what is wrong with it. People tell me topology is all about doing it and getting used to it but I just feel like I can never get the hang of topology... Please tell me why I've reached the false conclusion here!

Or, hang on since $\{a\} \in \tau$ does this mean $x-\{a\}=\{b,c\}$ is closed...so it's both open and closed? Thus $\{b,c\}$ can be the closure..?
 A: The closed sets are $\{\emptyset,X,\{a\},\{b,c\}\}$, so the intersection of the closed sets containing $A$ is $\{b,c\}$.
The fact that $\{b,c\}$ is also open is irrelevant: nowhere it is required that a closed set is not open. Subsets of $X$ can be


*

*open and closed

*open and not closed

*not open and closed

*neither open nor closed


The subsets of your three point space are


*

*$\emptyset$, open and closed

*$\{a\}$, open and closed

*$\{b\}$, neither open nor closed (closure is $\{b,c\}$)

*$\{c\}$, neither open nor closed (closure is $\{b,c\}$)

*$\{b,c\}$, open and closed

*$\{a,c\}$, neither open nor closed (closure is $X$)

*$\{a,b\}$, neither open nor closed (closure is $X$)

*$X$, open and closed


One can also look at neighborhoods. We see that $\{a\}$ is a neighborhood of $a$ which doesn't meet $A$. Therefore $a\notin\operatorname{Cl}(A)$.
The neighborhoods of $b$ are $\{b,c\}$ and $X$, because these are the only subsets of $X$ that contain an open set containing $b$. Since every neighborhood of $b$ meets $A$, we conclude that $b\in\operatorname{Cl}(A)$. Obviously $c\in\operatorname{Cl}(A)$, as $c\in A$.

In the standard topology of the reals, it's easy to make examples of sets which are in classes 2, 3 and 4. The only subsets of the reals that are both open and closed are $\emptyset$ and $\mathbb{R}$, because it is a connected space, but it's not so relevant. Any non connected space will provide examples of non trivial subsets that are both open and closed.
