Why is {$0$} closed set in the space of symmetric matrices I am preparing for my topology course and while going through some general ideas surrounding open sets of topological spaces, I read a statement written that suggested

{$0$} is a closed set in the space of symmetric matrices

To prove this statement, I just need a little clarifications surrounding the open sets of symmetric matrices:

*

*What kind of topology is defined on the set of symmetric spaces? I did some research honestly didn't understand much, it said something about having open balls as open sets in such spaces

*When dealing with spaces of matrices in general, how can i prove that a set is open, if the topology isn't clearly defined ( I heard my linear algebra teacher saying once we don't need a topology to know if a set is open or closed") not sure how is true ?

*How will be the open sets of space of symmetric matrices of n×n?

Now  if i knew what are the open sets, I could have explained why {$0$} is a closed set by just taking the complement of this set and showing its open in symmetric matrices. As far as my knowledge is concerned i am still new to many notations in topology, so I was hoping if i can get examples for simples dimensions when n=2 or 3?
Thankyou
 A: In general we cannot say whether a set is open/cloded without knowing the topology. However, most of the time we implicitely assume a certain topology without specifing. In basic calculus/linear algebra classes we usually assume that the topology for matrices is given by some norm (that is a function that measures the "seize" of this element). As the space of matrices is finite-dimensional all norms give the same topology (which is why we are not terribly concerned about which norm we actually pick).
The easiest way to think of the matrices as topological space (with topology induced by some norm) is by identifying it with some euclidean space. For example the $2\times 2$-matrices can be identifies with $\mathbb{R}^4$ via the linear map $(a_{ij})_{1\leq i,j\leq 2} \mapsto (a_{11}, a_{12}, a_{21}, a_{22})$.
Accordingly a set $U$ in the $2\times 2$-matrices is open iff for every $(a_{ij})_{1\leq i,j\leq 2}$ there exists $\varepsilon>0$ such that if
$$ \vert a_{ij} - b_{ij}\vert <\varepsilon,$$
then $(b_{ij})_{1\leq i,j\leq 2}$ belongs to $U$. Which is the same as we would do in $\mathbb{R}^4$.
If you want to define open sets in the symmetric matrices, you can use the same definition where you ask in addition $(a_{ij})_{1\leq i,j\leq 2}, (b_{ij})_{1\leq i,j\leq 2}$ to be symmetric matrices.
All of the above works just fine if we replace $2$ by any positive integer. Are you able to show with those definitions that $\{ 0\}$ is a closed set?
In general, one would introduce the concept of the subspace topology, but I think that would confuse you more than it would help you. However, once you have digested the things above you might want to look it up.
