Why do these conditions ensure symmetric positive-definiteness? Forgive me, if I have made a blunder or missed something obvious, I'm not a mathematician! 
I'm trying to understand 2 seemingly simple lines of maths - and understand how the conclusions are drawn from these. So here are the three lines (or rather two lines of maths, one statement at the end):
Let:
$L(n) := \{ A \in M(n) : A^T = A   \}$
$P(n) := \{ A \in L(n) : A > 0 \} $
"The boundary of $P(n)$ is the set of singular positive semidefinite matrices"
So from what I understand here, the set L contains symmetric ($A^T = A$) $n$-dimensional square matrices with real entries ($A \in M(n)$). Then I have presumed that the part, $A > 0$, refers to the determinant of $A$, and not $A$ itself. 
I am also inferring (possibly incorrectly) that $P(n)$ is, therefore, the space of SPD matrices..?
So why does ensuring we have square, real, symmetric matrices whose determinant is above 0 (this means they're invertible..?) ensure that the eigenvalues of said matrices are above 0 (my understanding of what makes a matrix positive definite)?
 A: As others have stated, $P(n)$ is intended to represent the set of all positive definite $n \times n$ matrices (not the set of all matrices with positive determinant, as bidby initially thought). We let $A \succ 0$ denote a positive definite matrix and $A \succeq 0$ denote a positive semidefinite matrix.
Claim: $\partial P(n) \overset{\textrm{Def}}{=} \overline{P(n)} \setminus \mathring{P(n)} = \{A \in \mathbb{R}^{n \times n} | A \succeq 0, \nexists A^{-1}\}$.
Proof: Let $A \in \partial P(n)$, $(A_k)_{k \in \mathbb{N}} \subseteq P(n)$, and $z \in \mathbb{R}^n$. Then
$$
z^T A z = z^T \left( \lim_{k \to \infty} A_k \right) z = \lim_{k\to\infty} \underbrace{\left( z^T A_k z \right)}_{\geq 0} \geq 0,
$$
since the functions $f : \mathbb{R}^n \to \mathbb{R}, \; x \mapsto z^T x$ and $g : \mathbb{R}^{n \times n} \to \mathbb{R}^n, \; B \mapsto B z^T$ are linear (and linear transformations between finite dimensional vector spaces are continuous). Therefore, $A$ is positive semidefinite.
We now show that $A$ is singular. Since $A$ is positive semidefinite, all its eigenvalues are non-negative. We know that $A$ is not positive definite, since $\mathring{P(n)} = P(n)$ by this post. But if the eigenvalues were all positive, then $A$ would be positive definite. Hence, $A$ has $0$ as an eigenvalue -- i.e. $\exists$ eigenvector $v \in \mathbb{R}^n \setminus \{0\}$ with $Av = 0 \cdot v = 0$. Therefore, $\textrm{Ker}(A) \neq \{0\}$, so $A$ is singular (see 3. here).
(A trivial example of such an $A$ and such a sequence $(A_n)$ is $A := 0 \in \mathbb{R}^{1 \times 1}$ and $A_n := 1/n$.)
