Is the set $\mathcal{A}$ of all matrices whose trace is $0$ are nowhere dense in $M_n(\Bbb{R}),n≥2$ ?
This question is already asked yesterday in this site( see this ) and I already answered that question (see my answer in that post). Why I'm posting again here is to verify my answer.
Can anyone explain me 'what I'm doing wrong in that answer" ? because of down votes but I hope it is correct!
Here's that answer for reference:
Result: If $A$ is a subset of a metric space $(X,d)$, then $A$ is nowhere dense in $X$ if and only if $\overline{A}^c$ is dense in $X$
[For a proof, see this]
Note that $\mathcal{A}$ is closed
So, In order to prove $\mathcal{A}$ is nowhere dense in $M_n(\Bbb{R})$, we prove $\overline{\mathcal{A}}^c=\mathcal{A}^c=M_n(\Bbb{R}) \setminus \mathcal{A}$ is dense in $M_n(\Bbb{R})$
To prove $M_n(\Bbb{R}) \setminus \mathcal{A}$ is dense in $M_n(\Bbb{R})$, we prove every point of $M_n(\Bbb{R})$ is either a point of $M_n(\Bbb{R}) \setminus \mathcal{A}$ or a limit point of $M_n(\Bbb{R}) \setminus \mathcal{A}$.
Suppose $B \in M_n(\Bbb{R}) \setminus \mathcal{A}$ ,then we are done! So asume $B \notin M_n(\Bbb{R}) \setminus \mathcal{A}$. That is $B \in \mathcal{A}$. In this case we prove $B$ is a limit point of $M_n(\Bbb{R}) \setminus \mathcal{A}$
take $B=\begin{pmatrix}a_{11} & a_{12} &\dots&a_{1n} \\ a_{21} & a_{22} \ &\dots&a_{2n} \\ \vdots \\a_{n1} & a_{n2} &\dots&a_{nn}\end{pmatrix}$ with $a_{11}+\dots+a_{nn}=0$
Then consider the sequence of elements of $M_n(\Bbb{R})\setminus \mathcal{A}$ $$A_k=\begin{pmatrix}a_{11}+1/k & a_{12} &\dots&a_{1n} \\ a_{21} & a_{22}+1/k \ &\dots&a_{2n} \\ \vdots \\a_{n1} & a_{n2} &\dots&a_{nn}+1/k\end{pmatrix}$$ Then $A_k \rightarrow B$ and so $B$ is a limit point!
Please help!
Your answer is perfectly alright. I wouldn't add or change anything, there doesn't appear to be anything to clarify further in the presentation. If a comment is absolutely necessary, I would just note that adding $1/k$ to only the $(1,1)$ position also suffices.