Can a set be closed in one topology but neither open nor closed in another? Can a set be closed in one topology but neither open nor closed in another?
If we say that the complement of a open set is a closed set, i.e. if $S \subseteq X$ is open then $X \setminus S$ is closed, I argue that depending on the topology, it could actually be the case. Consider the Euclidean topology on $\mathbb{R}$. Clearly any closed interval $[a,b]$ with $a,b \in \mathbb{R}$ is a closed set in this topology.
However, given the trivial topology on $\mathbb{R}$, any interval $[a,b]$ is neither closed or open.
But according to this Wikipedia page, an equivalent definition of a closed set is that "a set is closed if and only if it contains all of its limit points". And from this point of view, the inverval $[a,b]$ is clearly closed.
So i guess that my question is: How can two (supposedly) equivalent definitios of a closed sets, have a set be closed in one and not the other?

The definition of a limit point depends on the topology too. 
Recall the definition of a limit point:
A limit point of a set $S$ in a topological space $X$ is a point $x$ (which is in $X$, but not necessarily in $S$) that can be "approximated" by points of $S$ in the sense that every neighbourhood of $x$ with respect to the topology on $X$ also contains a point of $S$ other than $x$ itself.
With this definition, and the trivial topology on $\mathbb{R}$, any point of $\mathbb{R}$ is a limit point of $[a, b]$.
