# Does the Banach-Alaoglu theorem imply finite dimensionalness?

Given a normed separable vector space $X$, the Banach-Alaoglu theorem states that the closed unit ball in the dual space $X^*$ is compact in the weak-* topology. Since $X$ is separable, that topology is metrizable.

Now, the Alaoglu theorem states that for any normed vector space, the closed unit ball is compact iff the space is finite dimensional. Applying this to $X^*$ allegedly yields that $X^*$ is finite dimensional (obviously false).

Where is my mistake?

• If $X$ is not finite-dimensional then the weak$^*$ topology on $X^*$ is strictly weaker than the norm-induced topology of $X^*.$ – DanielWainfleet Apr 17 '17 at 7:50
• the closed unit ball is compact iff the space is finite dimensional: Riesz theorem not Banach-Alaoglu – Zbigniew Aug 22 at 10:15
• Finite dimensional is the trivial cases. You do not see anything. That is the space is trivially separable and the weak-* topology coincide with the norm topology and then the space is separable, the unite ball is compact. – Zbigniew Aug 22 at 10:20

• You mean a norm topology? Or am I missing why should there be a unique norm on that space (up to equivalence)? Also, isn't the metric induced by the norm $\|x\|=d(x,0)$? – BOS Apr 16 '17 at 23:18
• If $d$ is a metric corresponding to the weak-* topology, $d(x,0)$ is not a norm. One reason it can't be a norm: every weak-* neighbourhood of $0$ contains infinite rays from the origin. – Robert Israel Apr 17 '17 at 2:27
• No, it does not induce the weak * topology. $\{\psi: \|\psi\| < 1\}$ is not a weak-* neighbourhood of $0$. – Robert Israel Apr 19 '17 at 6:57