Suppose that $X$ is a topological vector space where the topology is given by a metric $d$ on $X$. Assuming that the unit ball $$ B(0, 1) := \{x \in X : d(0, x) < 1\} \neq X, $$ is it necessarily true that $2$ times the unit ball $$ 2 B(0, 1) := \{2 x : x \in B(0, 1)\} $$ is bounded with respect to the metric $d$? Here we do not assume that $X$ is finite dimensional or that $d$ is translation invariant.
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1$\begingroup$ For any translation invariant metric, such as the one suggested, $2 B(0, 1)$ is contained in $B(0, 2)$ and therefore is bounded too. $\endgroup$– Chandan BiswasCommented Aug 18 at 17:36
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$\begingroup$ (I deleted a wrong example from a previous comment. Obviously there's no example with $X$ finite-dimensional, since closed balls are then compact.) A non-complete normed space should give an example, using a point $u$ in the completion, not in $X$, and finding a metric making this point $u$ be "at infinity". $\endgroup$– YCorCommented Aug 18 at 21:49
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$\begingroup$ First, even in finite dimension, lets say in $\mathbb{R}$, it is not a priori guaranteed that the unit ball with respect to the metric $d$ is convex, thus there is a possibility that the unit ball itself might not be bounded with respect to the usual $l^2$-norm. Secondly, could you please elaborate on how do you define the metric $d$ by adding one such point at infinity? Thanks. $\endgroup$– Chandan BiswasCommented Aug 19 at 7:58
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$\begingroup$ You're right about finite dimension. This indeed yields an example: let $d$, be the largest pseudodistance on $\mathbf{R}$ that is $\le$ the usual distance and such that $d(0,3^n)\le 1$ for all $n\ge 0$. Then $d$ defines the usual topology; twice the unit ball contains $\{2.3^n:n\ge 0\}$ which is unbounded ($d(0,2.3^n)=1+3^n$.) $\endgroup$– YCorCommented Aug 19 at 10:21
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$\begingroup$ Aha, so it’s like the edge of a flower with infinitely many increasingly and arbitrarily large petals. Thanks. $\endgroup$– Chandan BiswasCommented Aug 19 at 11:36
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