Belongs to / subset problem in topology I'm studying topology and I came across with the following problem which is a bit confusing to me: 

The set $\Large\tau$ is called a discrete topology if it is the
  collection of all subsets of non-empty set $X$.
If $X=\{a,b,c,d,e,f\}$ and $\Large\tau$ is the discrete topology on
  $X$, which of the following statements are true: 
(a) $X\in \Large\tau\;\;\;$   (b) $\{X\}\in\Large\tau\;\;\;$ (c)
  $\{\varnothing\}\in\Large\tau\;\;\;$ (d)
  $\varnothing\in\Large\tau\;\;\;$
(e) $\varnothing\in X\;\;\;$   (f) $\{\varnothing\}\in X\;\;\;$ (g)
  $\{a\}\in\Large\tau\;\;\;$ (h) $a\in\Large\tau\;\;\;$
(i) $\varnothing\subseteq X\;\;\;$   (j) $\{a\}\in X\;\;\;$ (k)
  $\{\varnothing\}\subseteq X\;\;\;$ (l) $a\in X\;\;\;$
(m) $X\subseteq \Large\tau\;\;\;$   (n)
  $\{a\}\subseteq\Large\tau\;\;\;$ (o) $\{X\}\subseteq\Large\tau\;\;\;$
  (p) $a\subseteq\Large\tau\;\;\;$
Hint: Precisely six of the above are true.

If I were to do this task I would say that (a), (d), (g), (i), (l), (n) were true. I wonder if I got it right or wrong? 
The problem can be found here Page 26 task 3
 A: (a) $X \in \mathcal{T}$  true, all subsets of $X$ are in the discrete topology on $X$ (and $X$ is of course in any topology on $X$ by the first axiom of topology as well).
(b) $\{X\} \in \mathcal{T}$: $\{X\}$ is not a subset of $X$, as $X \notin X$, so this cannot be true. (assuming the axiom of foundation)
(c) $\{\emptyset\}$ is also not a subset of $X$ (because otherwise $\emptyset \in X$ which is false, assuming, as I suppose we must assume here, that all of $a$ to $f$ are different from $\emptyset$ (a priori, $\emptyset$ can be an element of a set, but as $X$ is enumerated as $\{a,\ldots,f\}$, we assume this is not the case here (the question is somewhat badly formulated, IMHO). 
(d) $\emptyset$ is a member of any topology on $X$, so true.
(e) $\emptyset \in X$ we assume is false (see (c)).
(f) $\{\emptyset\} \in X$ we likewise assume to be false (see (c), we assume $a$ to $f$ are "unrelated" elements, not the empty set, or this set, e.g.)
(g) $\{a\} \in \mathcal{T}$ is true, because $\{a\} \subset X$ (because $a \in X$, by enumeration) and all subsets of $X$ are in $\mathcal{T}$.
(h) $a \in \mathcal{T}$ is false, as $a$ is not a subset of $X$ (we assume; a priori $a = \{b\}$ could hold, and then $a$ would be a subset of $X$!; this is part of the bad formulation criticism).
(i) $\emptyset \subseteq \mathcal{T}$ is true: all elements of $\emptyset$ are in $\mathcal{T}$, this is an empty statement as there are no elements. Or equivalently, $\emptyset \subseteq A$ is true whatever $A$ is.
(j) $\{a\} \in X$ is false (we assume, see before, as $b$ could equal $\{a\}$ for all we know... ).
(k)  $\{\emptyset\} \subseteq \mathcal{T}$ is true, as $\emptyset \in \mathcal{T}$ (inclusion: every element left is an element right..)
(l) $a \in X$, true by inspection.
(m) $X \subseteq \mathcal{T}$, false, as $a \in X$ but not $a \in \mathcal{T}$, see (h).
(n) $\{a\} \subseteq \mathcal{T}$ is false, as $a \in \{a\}$ but not $a \in \mathcal{T}$, see (h) again.
(o) $\{X\} \subseteq \mathcal{T}$ is true, as the only element left, $X$, is indeed an element of $\mathcal{T}$, see (a).
(p) $a \subseteq \mathcal{T}$ I call false (it's not known what the elements of $a$ are, if $a = \{b\}$ happened to be true, then it would hold, so really we cannot say based on what's given).
Under the given assumptions, indeed 6 are true.
