Given $a_1,\ldots,a_n,b\in\Bbb R$, prove $S=\{x\in\Bbb R^n\mid \langle a,x\rangle>b\}$ is open via definiton

Question

Let $a_1,\ldots, a_n,b\in\Bbb R.$ Prove the set $S=\{(x_1,\ldots,x_n)\in\Bbb R^n\mid a_1x_1+\cdots+a_nx_n>b\}$ is open.

I would like to prove the claim via definition, that is $\forall x\in S,\exists r>0$ such that $B(x,r)\subseteq S.$ I know, if we rewrite $S=\{x\in\Bbb R^n\mid\langle a,x\rangle>b\}$ like here,since $T_a:\Bbb R^n\to\Bbb R,T_a(x)=\langle a,x\rangle$ is (a) continuous (linear functional), $S=T_a^{-1}((b,+\infty)),$ but I want to avoid multivariate functions and continuity at the moment. It appeared to me Cauchy-Schwarz could be used at some point, but I don't have any information about the scalars $a_1,\ldots,a_n,b\in\Bbb R.$ I fell into trap of defining a sequence $b_k=a_1x_k^1+\cdots+a_nx_k^n$ in $\Bbb R$ that would converge to $b$ using $b_k>b,\forall k\in\Bbb N\lim\limits_{k\to\infty}b_k\ge b,$ but that could only show us, if $x_k\to x,$ then $x$ might be a limit point not in $S$.

How can we show, $\forall x\in S,\exists r>0$ such that $\forall y\in\Bbb R^n,d(x,y)<r\implies \langle a,y\rangle>b$?

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Edit: I think this might have been the answer I was looking for that can be generalized to higher dimensions.

Answer 1

You could employ a sequential approach to show that $S^c=\{\vec{v}\in \mathbb{R}^n:\vec{v}\cdot \vec{a} \leq b\}$ is closed. Here, $\vec{a}=(a_1,...,a_n)$. To avoid any triviality, assume $\vec{a}\neq \vec{0}$.

Let $\{\vec{x}_k\}_{k=1}^{\infty}$ be a sequence of points in $S^c$ that converge to some $\vec{x}\in \mathbb{R}^n$. Pick $\epsilon >0$. Find $K>0$ such that $k\geq K$ implies $\|\vec{x}_k-\vec{x}\|<\frac{\epsilon}{\|\vec{a}\|}$. By Cauchy Swartz, we have for $k\geq K$ that $$|\vec{x}_k \cdot \vec{a}-\vec{x}\cdot \vec{a}|=|(\vec{x}_k-\vec{x}) \cdot \vec{a}|\leq \|\vec{x}_k-\vec{x}\|\|\vec{a}\|<\frac{\epsilon}{\|a\|}\times \|a\| =\epsilon$$ In other words, $\big\{\vec{x}_k \cdot \vec{a}\big\}_{k=1}^{\infty} \longrightarrow \vec{x}\cdot \vec{a}$. Now, if $\vec{x}\cdot \vec{a}$ were larger than $b$, we could certainly find $J>0$ so that $j\geq J$ implies $|\vec{x}_j \cdot \vec{a}-\vec{x}\cdot \vec{a}|<\vec{x}\cdot \vec{a}-b$. This would mean $\vec{x}_J \cdot \vec{a}>b$ i.e. $\vec{x}_J\notin S^c$ a contradiction. So $\vec{x}\cdot \vec{a}\leq b \implies \vec{x}\in S^c \implies S^c$ is closed.

Answer 2

This is a problem with a linear algebraic definition, so how about using linear algebra to solve it?

The set $S$ is a (translated, rotated) half-space. So if $x\in S$, you could calculate the distance from $x$ to the plane $\{y\ |\ \langle a, y\rangle = b\}$ and use that to define $r$.

To calculate the distance, orthogonally project $x$ onto the span of $a$: $\bar{x} = \langle x, a\rangle a$. Then $r = |\bar{x} - ba|$.

Answer 3

With the Cartesian norm $\|x\|=(\sum_i|x_i|^2)^{1/2}$ we have the identity $\|a\|^2\|x\|^2-<a,x>^2=\sum \sum_{1\le i\le j\le n}(a_ix_j-a_jx_i)^2\ge 0,$ whence the Cauchy-Schwarz Inequality $|<a,x>|\le \|a\|\cdot \|x\|.$

Let $r_x=<a,x>-b.$

If $x\in S$ then $r_x>0$ so take $t_x>0$ such that $\|a\|t_x<r_x.$

Now if $x\in S$ then $B(x,t_x)=\{y\in \Bbb R^n:\|x-y\|<t_x\}$ is open by definition.

If $x\in S$ then $B(x,t_x)\subset S$ because $$\|x-y\|<t_x\implies$$ $$\implies <a,y>\;=\;<a,x>-<a,x-y>\;\ge$$ $$\ge\; <a,x>-|<a,x-y>|\;\ge$$ $$\ge\; <a,x>-\|a\|\cdot \|x-y\| \ge$$ $$ \ge\;<a,x>-\|a\|t_x\;>$$ $$>\;<a,x>-r_x=b.$$ And if $x\in S$ then $x\in B(x,t_x).$ Therefore $$S=\cup_{x\in S}\{x\}\subset \cup_{x\in S}B(x,t_x)\subset S.$$ So $S=\cup_{x\in S}B(x,t_x)$ is the union of a family of open sets, so $S$ is open.

Answer 4

I know OP asked for a direct way, but I also offer an alternative just so you see other ways to prove it. We can definte the function $f:\Bbb{R}^n\to\Bbb{R}, f(x_1,...,x_n)=\sum_{i=1}^{n}a_ix_i$. I leave it to you to show that $f$ is continuous. So, $S=f^{-1}((b,\infty))$ (can you see why?), and the preimage of an open set under a continuous function is open.