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Let $X$ be a normed space and let ball $X$ be a closed unit ball of $X$. What the relation between $X$ and ball $X$? If $Y$ is a normed space with ball $Y$=ball $X$. Is $X=Y$? How? Thank you!

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    $\begingroup$ Why do you keep calling it "ball $X$"? Wouldn't say $B$ be clearer? $\endgroup$ – carmichael561 Aug 19 '17 at 1:44
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Let $B_X$ denote the closed unit ball of $X$. Here's a few well-known results regarding the unit ball.

1) $X$ is finite dimensional iff $B_X$ is compact.

2) $B_{X^*}$ is weak* compact.

3) $X$ reflexive iff $B_X$ is weakly compact.

4) $X$ separable iff $B_{X^*}$ is weak* metrizable.

For your second question, consider the following.

If $X$ and $Y$ are subspaces of some normed space such that $B_X\subseteq B_Y$, then $X\subseteq Y$.

Proof: Fix $x\in X$. If $x=0$, then $x\in Y$. Otherwise, $x/\|x\|\in B_X\subseteq B_Y$, so we deduce that $x=\|x\|(x/\|x\|)\in Y$. $\square$

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  • $\begingroup$ Can I use this theorem in Rudin's book to show $X=Y$? Let $X$ be a TVS, let $U$ be a neighborhood of $0$, and let $0<r_1<\cdots<r_n$ and $r_n\rightarrow\infty$ as $n\rightarrow\infty$. Then $X=\bigcup\limits_{n=1}^{\infty}r_nU$. $\endgroup$ – Answer Lee Aug 19 '17 at 18:30
  • $\begingroup$ @AnswerLee Yes that is essentially the idea. $\endgroup$ – John Griffin Dec 8 '17 at 3:49

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