Sequences in a subset of $\mathbb R^4$ I'm having some problems with the following question:
Consider $X=\{x \in \mathbb R^4 \mid x_1x_4 - x_2x_3\neq 0\}$. Show that given $a \in \mathbb R^4$ exists a sequence $(x_n)_{n \in \mathbb N}$ with elements in $X$, such that $x_n \rightarrow a$.
My attempt was to show that $X$ is dense at $\mathbb R^4$ (it seemed more natural to me). So, the idea was to take an arbitrary $x$ in $\mathbb R^4$ and an open ball $B_x$ centered in $x$ with radius $\varepsilon$. Soon after I was trying to show that $B_x \cap X \neq \emptyset$. However, the solution did not come out, I was unable to define a general element that belongs to $X$ and $B_x$. Could someone give me a tip?
Thank you very much in advance.
 A: This is obviously true for any point $x\in X$ (take the constant sequence $x_n = x$). If $x \notin X,$ then it is on the zero set of $f(x) = x_1 x_4 - x_2 x_3 .$ The gradient of $f$ at $x_1, x_2, x_3, x_4$ equals $(x_4, -x_3, -x_2, x_1),$ which is a non-zero vector except at the origin. At the points $y$ other than the origin, your sequence is $y_n =  y + \frac{\nabla f}{n}.$ At the origin. the sequence $y_n = (1/n, 1/n, 1/n, 1/n)$ works.
A: The given problem is a very special caseof the fact that for all quadratic matrices $ A$ of dimension $m$ we have $\det(A-\lambda E)=0 $ for at most $m$ values, the eigenvalues of $A$, of $\lambda$. This is implied by the fact that $\det(A-\lambda E)$ is a polynomial of degree $m$ in $\lambda$.
A: Here's two ways. The first is elementary (and constructive!), the second is fun.

Let $a \in \mathbb{R}$. Then the point $(a,0,0,1) \in \mathbb{R}^4$ yields the $a$ you desire.
The sequence you want is just:
$$x_n = (a - a/(n+1), 0 , 0, 1)$$
Which always satisfies your requirement.
How did I come up with this? This is just the determinant of the matrix:
$$\begin{bmatrix} a - a/(n+1) & 0 \\ 0 & 1 \end{bmatrix}$$
and as the determinant is multilinear in its columns, I know $\text{det}$ of this matrix is $(a - a/(n+1))\text{det}(I_2)$, where $I_2$ is the identity matrix with determinant 1.
The above sequence constructively shows what you want, but this whole nonsense with the determinant hints at something which gives a more fundamental understanding of why this works.

Consider the space $GL_n$, which is the space of $n \times n$ real-valued matrices with nonzero determinant.
Fact: the determinant is continuous. There are a few ways to look at this, but a short way to see it is that the determinant is essentially a really fancy polynomial.
From this fact, we can get that $GL_n$ has two path-connected components, that is $GL_n^+$ and $GL_n^-$, the regions where the determinant is positive or negative, respectively. How do we define a path from one positive determinant to another? Just associate the start of the path with a matrix, let its entries vary continuously to a matrix with the endpoint determinant, and because $\text{det}$ is continuous, we get a path.
From this, we can say that there always exists a sequence to $a$.
If $a$ is positive, consider any path $\gamma$ to $a$ in $GL_n^+$, which may be readably parameterized by $t$ in $[0,1]$. Then define:
$$x_n = \gamma(1-1/n)$$
Each element of this sequence is a matrix with a given determinant, but that's essentially the same thing as what you want.
If $a$ is negative, consider any path $\gamma$ to $a$ in $GL_n^-$. The reason why we care about the sign of $a$ and distinguishing which sign of $GL_n$ we keep the path in is because we don't want the path to go through $0$ (why?)
