Does a topological space contain the closed sets? I am confused concerning the relation between closed sets and a topological space.
I know that a topological space is mostly defined in terms of open sets via the underlying topology and ist requiered properties. The sets defining the topology are then called open sets.
Given the de Morgan laws one could restate this in terms of closed sets. But this will then necessarily imply that the closed sets define the topology and thus the topological space which contains them. That the topological space will contain both open and closed sets makes me totally confused. 
I know that one of the differences between a sigma algebra and a topological space is precisely this: through the complementation the sigma algebra contains also the closed sets whereas the topological space does not.
Can somebody help me clarify this confusion ?
Thanks.
 A: Isn't this a variant of the old member/subset confusion?
A topological space is a set $X$ together with a topology $\tau$ being the set of open subsets of $X$. The topology contains open sets - that is the set themselves are members of the topology. The set $X$ on the other hand has open sets as subsets (but the open sets are them selves not members of $X$). Of course there are closed sets as well that are subsets of $X$ and at least some of them are members of $\tau$ as well (namely $X$ and $\emptyset$).
Since closedness is defined in terms of openness we could have defined topological spaces instead using a set $X$ together with a topology $\phi$ being the closed subset of $X$ - we would then have slightly different rules about $\phi$ than the normal definition. But we don't normally do this (but it could perhaps be an excercise to define topological space this way and then prove that it's essentially equivalent to the standard definition).
A: I don't understand your doubt. If $(X,\tau)$ is a topological space, then $\tau$ is (by definition) the set of open subsets of $X$. And each element of $\tau$ is a subset of $X$. On the other hand$$\{X\setminus A\,|\,A\in\tau\}\tag{1}$$is the set of closed subsets of $X$ (by definition of closed subset). And the elements of $(1)$ are also subsets of $X$. I hope that this helps.
A: $O$ is a open set if and only if $O^c$ is a closed set. We take complement relative to the topological space, for example, if $X$ is the topological space, and $G\subset X$, you consider $G^c=X\setminus G$. This is, yes, the closed sets lies in the topological space, NOT (in general) in the topology.
Being rigours, a topological space $(X,\tau)$ (generally it is used to say that $X$ is the topological space, forgetting tau) is a set $X$ with a subset of the power set of $X$, namely $\tau$ with certain properties. And you say that the elements of $\tau$ are open sets of $X$. 
Example: Let $X=\{0,1\}$, and $\tau=\{\emptyset, \{0\},X\}$, then we have that $\tau$ define a topology on $X$ (prove it). Note that que closed sets in this case are $\{\emptyset,\{1\},X\}$.
A: Ah, I think I see your problem:

"I know that a topological space is mostly defined in terms of open
  sets via the underlying topology"

The word "definition" and "defined" is ambiguous.  Here is an anology:
A triangle is defined by its three sides.
We may define a triangle by one side and the two angles adjacent to it.
Those sentences are both true but seem contradictory.
A triangle has $!6!$ components three sides, three angles.  But if we have three of those components we can (usually) determine completely the entire triangle.
But we classify a triangle by its three sides.
The same it true of a topology.
A topological space is a set, and an assignment of all possible subsets to either open, closed, both or neither so that the assignment conforms to specific conditions.
We classify a topological space by a class of subsets that are stated to be open.  From that class we can determine completely and unambiguously, what status (open, closed, both, or neither) all subsets of the topological space are.
So we say $(G, \tau)$ is a topology where $G$ is a set, and $\tau$ is a list of open subsets.  
Okay, suppose we have a set $G$ and $\rho$, a list of closed subsets.  Can we create a topology from this?  Of course!  If $\rho$ is a list of closed subsets* then $\tau = \{L^c| L \in \rho\}$ will be the bases list of open subsets.
So the topology that we created with $G$ and $\rho$ is $(G, \tau =\{L^c| L \in \rho\})$.
There's absolutely nothing strange about this.  The topology was created by us knowing its closed subsets, but the topology is still a set and a class of open sets.  It's just that the open sets are the compliments of the closed sets.  Not the closed sets themselves.
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"The sets defining the topology are then called open sets."
That's sort of a weird and backward way of looking at it.  The topology is distinguished and recognized by which of its subsets are open.  But we can define the topology in any dang way we want--- so long as the definition necessarily means precisely those subsets are the subsets that are open.
So a set is closed if and only if it's complement is open.  So if we chose to define the topology by listing the closed sets we'd get the same result.  The topology's open sets can be determined by the compliment of the closed sets.
But we wouldn't call those defining sets open sets.  We'd call them closed sets because they were defined to be the closed sets.
"That the topological space will contain both open and closed sets makes me totally confused."  Topologies always contain both open and closed sets.  That's not a big deal.
Okay, let's do an example.  Let $X = \{1,2,3,4\}$.  It has $16$ subsets.  And let $\tau$ = $\{\{2\}, \{1,3\}\}$ that means the open sets will be $X, \emptyset, \{2\}, \{1,3\}, \{2\}\cup \{1,3\} = \{1,2,3\}$.  And the closed sets will be the complements of those $X, \emptyset, \{1,3,4\}, \{2,4\} ,\{4\}$. The remaining 8 sets, $\{1\},\{3\},\{1,2\}, \{1,4\},\{2,3\},\{3,4\}\{1,2,4\},\{2,3,4\}$ are neither open nor closed.
Now we could just as easily define the topology by listing the closed sets.  We'll call the base of closed sets $\rho$. So we can define a topology via $\rho = \{\{1,3,4\},\{2,4\}\}$.  From this we would conclude the closed sets are $X,\emptyset,\{1,3,4\},\{2,4\}, \{1,3,4\} \cap \{2,4\} = \{4\}$.  We'd take the complements to find the open sets.  $X, \emptyset, \{2\},\{1,3\},\{1,2,3\}$. The remaining 8 sets, $\{1\},\{3\},\{1,2\}, \{1,4\},\{2,3\},\{3,4\}\{1,2,4\},\{2,3,4\}$ are neither open nor closed.
This is the exact same topology as it has the exact same open, closed, both and neither subsets.
ADDENDUM
In light of precise definitions:
Let $X$ be a set:  Let $\tau$ be a list of subsets of $X$ so that: 1) $X, \emptyset \in \tau$.  ii) for any $U, V \in \tau$ then $U\cap V \in \tau$ , iii) for any $\phi \subset \tau$ then $\cup_{U \in \phi} U \subset \tau$.
We say that $\tau$ is a topology on $X$.
Now let $\rho$ be a list of subset of $X$ so that 1) $X, \emptyset \in \rho. ii) for any $U, V \in \rho $U\cup V \in \rho$ , iii) for any $\phi \subset \rho$ then $\cap_{U \in \phi} U \subset \rho$
We set and $\rho$ is  a notology on $X$.
Exercise:  It can be verified that $\tau$ is a topology on $X$ if and only if $\{U^c| U \in \tau\}$ is a notology on $X$ (and vice versa).
Now a topological space $\langle X, \tau\rangle$ is the set $X$ and a topology $\tau$ on it, with the understanding that the $U\in \tau$ have the property of being "open".  The $V \in \{U^c|U\in \tau\}$ have the property of being "closed".  And all other subsets have neither property.
ANd let's say  a notological space $\rangle X, \rho\langle$ is the set $X$ and a notology $\rho$ on it, with the understanding that the $U\in \rho$ have the property of being "closed".  The $V \in \{U^c|U\in \rho\}$ have the property of being "open".  And all other subsets have neither property.
Notice that notological space $\rangle X, \rho\langle$ and the topological space $\langle X, \{U^c|U \in \rho\} \rangle$ are the exact same construct!
That really is all the comment is saying.
