For a normed space, is a vector subspace of it with the restriction norm also a normed space? Question. Let $X = (X,\|\cdot\|)$ be a normed space and let $S \subset X$ be a subspace of $X$. Is $S$ a normed space (with the norm $\|\cdot\|$ of $X$) ?
My intuition. All we need to say for $S$ to be a normed space is that $S$ is a vector space and that $\|\cdot\|$ defines a norm on $S$. Since $S$ is a subspace of $X$, it also verifies the axioms to be a vector space and thus, $S$ is a vector space. On the other hand, if $\|\cdot\|$ is a norm on $X$ it will surely also be a norm on $S$. Thus, the answer is affirmative.
Is my thinking correct?
Thanks for any help in advance.
 A: Let $(X,\|.\|)$ be a real normed vector space and let $S\subset X$ a subspace of $X$.

*

*$S$ is a vector space(in particular $0\in S)$;

*Let $$\|.\|_S:S\to \Bbb R^+$$
$$x\mapsto \|x\|$$
Then
a) $\forall \lambda \in \Bbb R,\forall x\in S ,\|\lambda x\|_S=\|\lambda x\|=|\lambda|\| x\|=|\lambda|\| x\|_S$;
b) $\| 0\|_S=\| 0\|=0;$
c) $\forall x\in S, $ if $\| x\|_S=0$, then $\| x\|=0$, then $x=0$;
d) $\forall x,y \in S, \| x+y\|_S=\| x+y\|$ and $\| x+y\|\leq\| x\|+\| y\|$, i.e. $\| x+y\|_S\leq\| x\|_S+\| y\|_S$
Thus, $(S,\|.\|_S)$ is a real normed vector space.
A: If $X=B_1(0)$, the unit ball in $R^n$ under Euclidean norm, you can not take a linear subspace of it.
A topological subspace inherits the topology of the space.
Let $E\subset X$ be a subspace.
Then every open set $U\subset E$
is an intersection  $V\cap E$ where $V$ is an open subset of $X$.
Thus:
$(\forall x)(x\in U \implies x\in V\cap E\implies x\in V\implies (\exists B_r(x))(B_r(x)\subset V\cap E))$
Since $U=\cup_x{B_r(x)}$
$U$ is defined by the norm topology of $X$.
