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Consider a vector space $X$ equipped with a separating countable family of seminorms $(p_i)_{i\in\mathbb{N}}$. I'm told that the boundedness of a set in $X$ with respect to each one of the seminorms simultaneously is a stronger assumption than boundedness with respect to the metric defined by $d(x,y) = \Sigma \frac{1}{2^i}\min(1,p_i(x-y))$. So first thing I say to myself well this is a bit odd since then we have to choose which definition to use for boundedness. But then I realised that every metrisable locally convex space is bounded for that metric, unless I've missed something, isn't it ?

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Perhaps an example may help?

$X=\ell^\infty({\Bbb N})$. Let $(e_k)_{k\geq 1}$ be the canconical base. Then the family of semi-norms: $$ p_k(x)= |\langle e_k,x\rangle|, \ \ k\geq 1$$ gives rise to the weak topology on $X$. A sequence (of sequences) $\xi_n\in X$ goes to zero iff each coordinate goes to zero. A ball for the metric you wrote is very 'big', i.e. contains cylinder sets with constraints only on finitely many coordinates. Now, on the other hand if you look at a subset where each $p_k$ is bounded then that will not be open in the weak topology. If you impose even that each $p_k(x)<1$ then you will be looking at the unit ball in $X$ (which again is not open in the weak topology).

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