Are the following sets closed in certain topologies? Let $ X = R $ .  For the given set, determine whether or not the set is closed in: the usual topology, the half open topology, the half open line topology, the discrete topology, the indiscrete topology, and the finite complement topology.
I have a whole list.. I would just like some help for the first couple. 
$(1,5]$
Q
R - Q
Lastly, I was asked this same question for our tutorial last week but for open sets... I am wondering if there is a general way that these two questions are related? For example none of these are open in the indiscrete topology does that mean they are all closed in the indiscrete topology?
I guess for each of these I would have to go through and see if it satisfies each of the 3 conditions to be a topology? 
 A: Well, the first set is neither open nor closed in the usual topology.This is because since comp {(1,5]} = $(-\infty,1] \cup (5,+\infty) $ , which is a union of a half-open interval with an open interval in the usual topology on R, which is neither open nor closed. So the complement of (1,5] is neither open nor closed and so (1,5] is neither as well.  
By contrast, in the half-open topology-and I assume you mean the left half open topology here-the result here is rather interesting since the set is clopen i.e. both open and closed in R. This is It's easy to show the set of all half-open intervals is a base for a topology on R. But it's also clear that the complement of this is also open in the half open topology. This is because any open set in the usual topology is also open in the half open topology they can be expressed as a countably infinite union of half open intervals. I'll leave you to prove this yourself with the following hint: Consider an arbitrary open interval $(a,b) \subset$ R and the collection of half open intervals $ F_i\subseteq $(a,b) = { (a,$r_i$] | where $r_i \in Q$ and there exists $c\in R$ where a < $r_i$ < c $\leq b $ }. So this means the complement of this set is also open in the half open topology so (1,5] is both open and closed in this topology. 
In the discrete topology,clearly any set is open,so (1,5] is open. In the indiscrete topology, only the entire space and the empty set are open,so (1,5] is closed. The finite complement topology is the collection of open sets defined by either the empty set or any set who's complement is finite. Therefore since the complement of (1,5] in R  is not finite or empty, (1,5] is closed in this topology. 
Now let's consider Q as a subset of R. Consider the usual topology on R and consider any intervals bounded by rationals, ie. ($r_1$,$r_2$). Since a rational number lies in between any 2 real numbers and an irrational number lies in between any 2 rationals, there is no open interval in Q that lies entirely in Q. Similarly, none of it's complements lie entirely in Q, either. So Q is niether open nor closed in the usual topology. 
I don't have time to do the rest,sorry. But you can use the exact same reasoning  to determine the answers to your other problems. 
As for your last question, my first answer pretty much answered that-sets can be either both open and closed or neither in a given topology and we have to be really careful to ensure which it is. 
Good luck! 
A: In the following I will denote the irrational numbers by $I = \mathbb R \setminus \mathbb Q$.

First, let's consider $\mathbb R$ with the usual topology: 
A set in this topology is open if and only if for every point in the set there is an open ball around the point such that the ball is contained in the set. 
Applying this definition we see that any $\varepsilon$-ball around $5$ contains points outside $(1,5]$ hence this interval is not open. 
A set is closed if and only if its complement is open. So next we consider the complement $(1,5]^c = (-\infty,1] \cup (5,\infty)$. We note that any open ball around $1$ necessarily contains points outside $(1,5]^c $ hence $(1,5]^c $ is not open and therefore $(1,5]$ is not closed. 
To summarise our findings: $(1,5]$  is neither closed nor open. 
Next we consider $\mathbb Q$. Here we use that a set is closed if and only if it equals its closure: of course, $\overline{\mathbb Q} = \mathbb R$ hence $\mathbb Q$ is not closed. 
Let $q$ be any point in $ \mathbb Q$. Then we use the fact that for any two rationals $p < t$ there exists an irrational $i$ with $p < i < t$. Therefore every open ball around $q$ must contain points not in $\mathbb Q$ hence $\mathbb Q$ is not open.
So $\mathbb Q$ is neither open nor closed. 
Finally, you can use the argument we used for $\mathbb Q$ to reach the same conclusion about $I$.

Next we consider $\mathbb R$ with the lower limit topology (a.k.a. half open topology). It helps to recall that this topology has a basis consisting of all $[a,b)$ for all $a,b \in \mathbb R$.
It follows from the properties of a basis (union and intersection properties) that every open set in the half open topology is of the form $[a,b)$ (where $b$ can also be $\infty$). From this observation it becomes clear that any open set containing $5$ will contain points outside $(1,5]$ hence $(1,5]$ is not open. We can repeat this argument for the point $1$ when considering the complement to see that the complement is not open. Hence $(1,5]$  is not closed. 
We summarise: In the lower limit topology, $(1,5]$  is neither closed nor open.
Given any point $q$ in $\mathbb Q$ and any interval $[a,b)$ containing $q$ by using the same arguments as before we see that $[a,b)$ must contain an irrational number (between $q$ and $b$). Similarly, for the irrational numbers $I$ the interval contains a rational. Hence $\mathbb Q$ is neither closed nor open in the half open topology.

Unfortunately, I have never come across the "half open line topology" and doing a quick Google search did not come up with any useful hits. If you add a definition of this topology to your question I'm happy to help.

In the discrete topology every set is open.

In the discrete topology the only open (and therefore closed sets) are $\varnothing$ and $\mathbb R$. So in this case, all of the sets in question are neither open nor closed.

In the finite complement topology a set is open if and only if it is of the form $\mathbb R - \{x_1, \dots, x_m\}$. So again, all the sets in question are neither open nor closed in this topology. 
