Let $A \in \mathbf{R}^d \backslash \{0\}$ and $b\in\mathbf{R}$. Show that the hyperplane $H=\{x \in\mathbf{R} \;|\; : {A}^{T} x =b\}$ is closed in $\mathbf{R}^d$.
Asked
Active
Viewed 101 times
0
-
1Could you edit your question to clarify what you are asking? – mfl Oct 31 '14 at 13:59
-
I just edited the question and answered it – Olórin Oct 31 '14 at 14:09
-
Thanks for the edit, but please note I have changed one bit back to {A}^{T}. – Oct 31 '14 at 14:12
1 Answers
0
It is the inverse image of $\{b\}$, which is closed in $\mathbf{R}$, by the continuous map $x\mapsto {}^{t} A x $ from $\mathbf{R}^d$ to $\mathbf{R}$. It is therefore closed.
Please validate my answer if you are satisfied by it.
Olórin
- 12,040
-
-
$T$ or $t$ is the same, it means the transposed of the vector... Now, you can validate ;-) – Olórin Oct 31 '14 at 14:16
-
$T$ or $t$ is the same, it means the transposed of the vector, you can put it after of before $A$, it is just a matter of notation... – Olórin Oct 31 '14 at 14:22
-
Is there any other way to prove it? I don't understand the continuous map part, or why this implies it is closed. – Oct 31 '14 at 14:28
-
A map $f : X \rightarrow Y$ is continuous if and only if for every closed subset $Z$ of $Y$ the subset $f^{-1} (Z)$ of $X$ is closed in $X$. – Olórin Oct 31 '14 at 17:10
-
You know that $H$ is closed if and only it is stable by convergence : let $(x_n)n $ be a sequence of elements in $X$ converging to $x\in\mathbf{R}^d$. Then for each $n$ you have ${}^{t} A x = {}^{t} A (x-x_n) + \underbrace{{}^{t} A x_n}{=0}$ because $x_n \in H$, and then $|{}^{t} A x| = | {}^{t} A (x-x_n) | \leq |A| |x-x_n| \rightarrow 0$ as $n\rightarrow +\infty$. Passing to the limit, you get $|{}^{t} A x- = 0$, that is ${}^{t} A x = 0$, showing that $x\in H$. Therefore, $H$ is closed. – Olórin Oct 31 '14 at 17:14