Is the set of diagonalizable, complex matrices open in the set of square matrices? 
Is the set of complex, diagonalizable matrices open in the set of square matrices?

I asked myself this question and I tried to prove it somehow. However, I don't have any good approach so far.
What is the best way to prove this?
 A: No, the set of diagonalizable matrices is not open:        
the matrix    $  \begin{pmatrix}    1 &\epsilon\\    0 & 1            \end{pmatrix}        $ is diagonalizable only if $\epsilon=0$ .
[Reason: if such a matrix is diagonalizable its diagonalization must be $I$, the unit matrix, because the only eigenvalue is $1$  and the matrix $I$ is similar only to itself]  
Edit: arbitrary fields
The above is meant implicitly  over  $\mathbb R$ or $\mathbb C$ with their usual metric topology, but more algebraically minded users may take as ground field an arbitrary field with its Zariski(=cofinite) topology.
A: Take an arbitrary open ball around the identity matrix. This will contain matrices that have $1$ on the diagonal and $\epsilon$ on the second diagonal above the diagonal for $\epsilon > 0 $, if $\epsilon$ if sufficiently small. This is a scalar multiple of the Jordan normal form of a matrix that cannot be diagonalized (putting $1/\epsilon$ on the diagonal and $1$ on the second diagonal). So no, your set of diagonalizable matrices is not open.
A: An idea: assuming we're talking of $\;M_n(\Bbb C)\;$ embedded in $\;C^{n^2}\;$ and with the inherited topology there, let us look at
$$\forall\,n\in\Bbb N\;\;,\;\;\;A_n:=\begin{pmatrix}0&\frac1n&0&\ldots&0\\0&0&0&\ldots& 0\\\ldots&\ldots&\ldots&\ldots&\ldots\\0&0&0&\ldots&0\end{pmatrix}$$
Obviously $\;\{A_n\}_{n\in\Bbb N}\rlap\subset{\,/}\;Diag_n(\Bbb C)\;$, yet $\;A_n\xrightarrow[n\to\infty]{}0_n\in Diag_n(\Bbb C)\;$ , so...
