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!

  • $\begingroup$ There is an answer to this in the comments under your answer: '$\mathcal{A}$ is nowhere dense in $\mathcal{S}$' and '$\mathcal{A}^c$ is dense in $\mathcal{S}$' are not equivalent statements. Proving the latter does nothing towards proving the former. $\endgroup$ – Steven Stadnicki Sep 26 '18 at 2:50
  • $\begingroup$ No! Do you agree the result I mentioned in the answer? I use that result. In our case $A$ is closed so its closure equals A $\endgroup$ – Chinnapparaj R Sep 26 '18 at 2:54
  • 1
    $\begingroup$ You did nothing wrong. People keep making the same misunderstanding Steven made. They keep missing that you immediately stated that for closed sets, nowhere-denseness is equivalent to denseness of the complement. $\endgroup$ – Bob Jones Sep 26 '18 at 5:10
  • $\begingroup$ @BobJones: yes! I clearly mention $\overline{A}^c=A^c$ since it is closed. But why everyone do not accept that? What's wrong? Why there is a big list of arguments in the comments of that answer? why everyone down voting that? $\endgroup$ – Chinnapparaj R Sep 26 '18 at 5:14
  • 1
    $\begingroup$ Probably because they see you jumping into proving that $A^c$ is dense and they read this part and complain about it before they read the beginning. I dunno. $\endgroup$ – Bob Jones Sep 26 '18 at 5:27

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.

  • 1
    $\begingroup$ Btw, I've also upvoted and commented on your main answer. I hope the downvotes are reversed by the casters, should they return to the post. $\endgroup$ – Brahadeesh Sep 26 '18 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.