The openness of the set of positive definite square matrices

Let $\mathbb{R}^{n\times n}$ be the vector space of square matrices with real entries.
For each $A\in \mathbb{R}^{n\times n}$ we consider the norms given by:
$$\displaystyle\|A\|_1=\max_{1\leq j\leq n}\sum_{i=1}^{n}|a_{ij}|;$$
$$\displaystyle\|A\|_\infty=\max_{1\leq i\leq n}\sum_{j=1}^{n}|a_{ij}|;$$
$$\displaystyle\|A\|_\text{max}=\max\{|a_{ij}|\}.$$
Matrix $A\in \mathbb{R}^{n\times n}$ is said to be positive definite iff
$$\langle Ax, x\rangle> 0 \quad \forall x\in\mathbb{R}^n\setminus\{0\}.$$
Let $S$ be the set of all positive definite matrices on $\mathbb{R}^{n\times n}$. Prove that $S$ is an open set in $(X,\|.\|_1)$,
$(X,\|.\|_\infty$), $(X,\|.\|_\text{max})$.

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Solutions Collecting From Web of "The openness of the set of positive definite square matrices"

Restricting to the unit ball is always illustrating. Let $A$ be a given positive definite matrix, then there is $\delta>0$ such that
<Ax,x>\ge\delta
for all $\|x\|=1$.

We use the 2-norm, defined by
\|A\|=\operatorname{sup}_{\|x\|=1}\|Ax\|,
which is equivalent to any other norms.

If $B$ is very close to $A$, say, $\|B-A\|<\epsilon$, then

|<Bx,x>-<Ax,x>|=|<(B-A)x,x>|<\epsilon\|x\|^2,
so if you restrict to the unit ball again then you can bound $<Bx,x>$ from below using positive definiteness of $A$ and controlling $\epsilon$, and this will lead to the positive definiteness of $B$.