Prove that the set of $n \times n$ matrices with determinant $1$ is unbounded closed with empty interior in $\mathbb{R}^{n^{2}}$. 
Prove that the set of $n \times n$ matrices with determinant $1$ is unbounded closed with empty interior in $\mathbb{R}^{n^{2}}$.

The aplication $\det$ is continuous, so the inverse image of a closed set, is a closed set, right? Then, $\lbrace 1 \rbrace$ is closed set of $\mathbb{R}$ and the set of $n \times n$ matrices with determinant $1$ is $\det^{-1}(1)$, so it should be closed. Is correct?
For the others afirmations, I have no ideia how to start. Thanks for any hint!r
 A: The unboundedness can be seen by taking a diagonal matrix $D$ with $D_{11} = 1/N$ , $D_{22} = N$, and $D_{jj} = 1$ for $j = {3, \dots, n}$ for all $N \in \mathbb{R} \setminus \{0\}$.
For the empty interior, suppose $A$ were an interior point of the set 
$$SL(n) = \{A \in \mathcal{n \times n}: \text{det}(A) = 1\}.$$ 
Then we would have: there exists some $\delta > 0$,  for all $\varepsilon \in (-\delta, \delta)$
\begin{align}
  \tag{$\star$}
  \label{eq:1}
  A + \varepsilon E_n \in SL(n),
\end{align}
where $E_n$ is the all $1$ matrix. Let $p(\varepsilon) = \det(A + \varepsilon E_n)$. $p(\varepsilon)$ is a polynomial in $\varepsilon$ of degree $n$. But \eqref{eq:1} implies $p(\varepsilon) = 1$ has infinitely many solutions (every open interval has same cardinality as $\mathbb R$). This clearly contradicts fundamental theorem of algebra.
A: The set of all $n\times n$ real matrices with determinant 1 vied as a subset of $\mathbb{R}^{n^2}$ can be identified with the set
$$
E: = \{x\in \mathbb{R}^{n^2}: \ p(x) = 1 \},
$$
where $x\in \mathbb{R}^{n^2}$ is the "unwrapped" form of an $n\times n$ matrix (we put each row side by side), and $p(x)$ is the determinant, which is a polynomial in $x$.
Since $p$ is a continuous function, the set $E$ is closed. Next, for any $\varepsilon>0$ we have 
$$p(\varepsilon x_1,  x_2, ..., x_{n+1}, \frac {1}{\varepsilon}x_{n+2},x_{n+3},...,  x_{n^2}) = p(x_1,x_2,....,x_{n^2})$$
where $x$ corresponds to the identity matrix, which is in $E$ so, $E$ is not bounded (compare with the answer by @MartinArgerami above).
Finally, we are left to show that $E$ has no interior points. In fact, $E$ is a set of measure $0$, as a $0$-set of the non-zero polynomial $p(x) - 1$, and hence is free of interior points (it is a well-known fact, and not very hard to prove, that unless the polyomial is identically $0$, then it's $0$-set has measure $0$, see here for example).
A: To discuss boundedness, you need a metric. Now, since all norms on $M_n(\mathbb R)$ are equivalent, we may choose some norm that suits us. For instance, let us use 
$$
\|A\|=\max\{|A_{kj}|:\ k,j=1,\ldots,n\}. 
$$
Your set is unbounded because for instance you have for any $n\in\mathbb N$
$$
A=\begin{bmatrix} n&0&0&\cdots&0\\ 0&1/n&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots &\vdots&0&\ddots&\vdots\\ 0&0&0&\cdots&1,
\end{bmatrix}
$$
with $\|A\|=n$. 
As for the empty interior, we need to show that any $A\in\det^{-1}(\{1\})$ is arbitrarily close to a matrix not in the set. One way to do this would be to use the Schur Decomposition to write $A=VMV^T$ with $V$ orthogonal and $M$ upper triangular ($\det M=\det A$ and $\det M$ is the product of its diagonal entries). Now, given $\varepsilon>0$,  let $M'$ be equal to $M$ but with the $1,1$ entry added to $\varepsilon$. Then $A'=VM'V^T\not\in\det^{-1}(\{1\})$, and (with a very crude estimate, using that all entries of $V$ are in $[-1,1]$)
$$
\|A-A'\|=\|V(M-M')V^T\|\leq n^2\|M-M'\|=n^2\varepsilon.
$$
As $n$ is fixed, we can make $n^2\varepsilon$ as small as we want. So $\det^{-1}(\{1\})$ has empty interior. 
