Why is the determinant as a function from $M_n(\mathbb{R})$ to $\mathbb{R}$ continuous? Can anyone explain precisely and rigorously? So far, I know the explanation which comes from the facts that
Also I have the confusion regarding the metric on $M_n(\mathbb{R})$.
$M_n(\mathbb R)$ is just $\mathbb R^{n^2}$ with the euclidian metric.
determinant is continuous, because it is a polynomial in the coordinates $$ \det(X) = \det ([x_{i,j}])= \sum_\sigma \text{sgn}(\sigma) \prod_{i=1}^{n} x_{\sigma(i),i}, $$ where $\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$ is a permutation.
Recall that the determinant can be computed by a sum of determinants of minors, that is "sub"-matrices of smaller dimension.
Now we can prove by induction that $\det$ is continuous:
Suppose that for $n$ we have that $\det$ is continuous on $M_n(\mathbb R)$, let $A\in M_{n+1}(\mathbb R)$. We know that $\det A$ can be calculated as the alternating sum over one of first row, when calculating the $\det$ of the appropriate minor.
So $\det A$ is written as a sum and scalar multiplication of $\det$ on a smaller dimension. From the induction hypothesis these are continuous and therefore $\det$ is continuous on $n+1\times n+1$ matrices.
The coefficient maps $A\longmapsto a_{i,j}$ are continuous because they are linear on the finite-dimensional vector space $M_n(\mathbb{R})$. Here you want to refer to the topology of the latter as a normed space, which does not depend on the norm since they are all equivalent in finite dimension. Then the determinant is a polynomial in the coefficients, so it is continuous by composition of continuous maps.
This answer is for $\epsilon-\delta$ fans!
Consider determinant as a function of the form $\det: \mathcal{M}_n(\mathbb{R})\to\mathbb{R}$. What are the elements of its domain? Indeed, they are functions! More specifically, if $A\in\mathcal{M}_n(\mathbb{R})$ then $A$ is a function of the form $A:\{1,\dots,n\}\times\{1,\dots,n\} \to \mathbb{R}$ which we usually call a matrix. Furthermore, the domain and target space of $\det$, i.e. $\mathcal{M}_n(\mathbb{R})$ and $\mathbb{R}$, are finite dimensional normed vector spaces. To investigate the continuity of $\det$, we have to choose norms on $\mathcal{M}_n(\mathbb{R})$ and $\mathbb{R}$. However, these normed spaces are finite dimensional and it does not really matter what type of norm we choose as they are all equivalent in such spaces. Consequently, we shall choose the Frobenius norm $\lVert\cdot\rVert_F$ on $\mathcal{M}_n(\mathbb{R})$ and absolute value $|\cdot|$ on $\mathbb{R}$ for further investigation. Then, continuity of determinant at $A \in \mathcal{M}_n(\mathbb{R})$ reads as
\begin{equation} \forall\epsilon>0, \quad \exists\delta>0, \quad \lVert B-A \rVert_F < \delta \implies |\det B - \det A| <\epsilon. \end{equation}
The hard work is to obtain an upper bound for $|\det B - \det A|$ in terms of $\lVert B-A \rVert_F$. Fortunately, this can be done.
Lemma. Let $A, B \in \mathcal{M}_n(\mathbb{R})$. If $\lVert B-A \rVert_F \leq 1$ then we have \begin{equation} |\det B - \det A|\leq M \,\lVert B-A \rVert_F, \end{equation} where $M = n!\sum_{m=0}^{n-1}\,C_m^n \lVert A\rVert^{m}_F$ and $C_m^n$ is $n$ choose $m$.
Continuity of determinant follows imidiately from this lemma. More specifically, given any $\epsilon > 0$, some $0 < \delta \leq \min\{1, \frac{\epsilon}{M}\}$ would work.
Remark 1. There is an extension of the above lemma for multi-linear forms on normed vector spaces. Determinant can be considered as an n-linear form on the normed vector space $\mathbb{R}^n\times\dots\times\mathbb{R}^n$.
Remark 2. Note that the constant $M$ depends on $A$ so that determinant is not uniformly continuous. Indeed, this is the case for multi-linear forms on normed vector spaces.
Proof. Firstly, pay attention to some definitions and notation. We know that
\begin{equation} \det A = \sum_{\sigma\in \mathcal{S}_n} \text{sgn}(\sigma) A_{\sigma(1),1}\dots A_{\sigma(n),n}, \end{equation} where $\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$ is a permutation of $(1,\dots,n)$ and $\mathcal{S}_n$ is the set of all permutations. Also, let $H:=B-A$ be the change in $A$ and define the following index set \begin{equation} I_m:=\{ (i_1,\dots,i_n)\,|\,i_j\in\{0, 1\},\,i_1+\dots+i_n=m \}. \end{equation} This is no magic and will show up in finding the upper bound naturally but I just wanted to make the definition before getting into that. Now, let us dive into it.
\begin{align} |\det B - \det A| &= \Big|\sum_{\sigma\in\mathcal{S}_n}\text{sgn}(\sigma)\big(B_{\sigma(1),1} \dots B_{\sigma(n),n} - A_{\sigma(1),1} \dots A_{\sigma(n),n}\big)\Big| \\ &\leq \sum_{\sigma\in\mathcal{S}_n}\big|\text{sgn}(\sigma)\big|\big|B_{\sigma(1),1} \dots B_{\sigma(n),n} - A_{\sigma(1),1} \dots A_{\sigma(n),n}\big| \\ &= \sum_{\sigma\in\mathcal{S}_n}\big|(A_{\sigma(1),1}+H_{\sigma(1),1}) \dots (A_{\sigma(n),n}+H_{\sigma(n),n}) - A_{\sigma(1),1} \dots A_{\sigma(n),n}\big| \\ &= \sum_{\sigma\in\mathcal{S}_n}\Big|\sum_{m=0}^{n-1}\sum_{i\in I_m}\big(H_{\sigma(1), 1}^{1-i_1}A_{\sigma(1),1}^{i_1}\big)\dots\big(H_{\sigma(n), n}^{1-i_n}A_{\sigma(n),n}^{i_n}\big)\Big| \\ &\leq \sum_{\sigma\in\mathcal{S}_n}\sum_{m=0}^{n-1}\sum_{i\in I_m}\big|H_{\sigma(1), 1}^{1-i_1}\dots H_{\sigma(n),n}^{1-i_n}\big|\big|A_{\sigma(1), 1}^{i_1}\dots A_{\sigma(n),n}^{i_n}\big| \\ &\leq \sum_{\sigma\in\mathcal{S}_n}\sum_{m=0}^{n-1}\sum_{i\in I_m} \lVert H\rVert^{n-(i_1+\dots+i_n)}_F \lVert A\rVert^{i_1+\dots+i_n}_F \\ &= \sum_{\sigma\in\mathcal{S}_n}\sum_{m=0}^{n-1}|I_m|\lVert H\rVert^{n-m}_F \lVert A\rVert^{m}_F \\ &= \sum_{m=0}^{n-1} |\mathcal{S}_n| \,C_m^n \lVert H\rVert^{n-m}_F \lVert A\rVert^{m}_F \\ &=\lVert H \rVert_F \sum_{m=0}^{n-1} n!\,C_m^n \lVert H\rVert^{n-m-1}_F \lVert A\rVert^{m}_F \\ &\leq \lVert H \rVert_F \, \Big(n! \sum_{m=0}^{n-1}C_m^n \lVert A\rVert^{m}_F\Big), \end{align} where the last inequality holds provided that $\lVert H \rVert_F \leq 1$. This completes the proof.
It's continuous because it's computable as a function from $\mathbb{R}^{n^2}$ to $\mathbb{R}$.
