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Let $(X,\tau)$ be a locally convex topological vector space and denote the product space

$$X^{\infty}=X\times X\times X\cdots:=\big\{x=(x_i)_{i\geq 1}:~ x_i\in X\big\}$$

If we endow $X^{\infty}$ with the topology induced by projection, i.e. for $\{x^n\}_{n\geq 1}\subset X^{\infty}$ and $x\in X^{\infty}$ then

$$x^n\to x\Longleftrightarrow x^n_i\to x_i,~ \forall i\geq 1$$

My question is whether $X^{\infty}$ is a locally convex space? Maybe my question is naive, but please let me know if someone knows the result. Many thanks!

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I edited the title into the form of a question. In the future, please try to put (a short version of) the question in the title, rather than titling posts "question about ...". – Theo Johnson-Freyd Jan 28 '14 at 4:27

Yes. A vector space is locally convex iff its topology can be defined by a family of semi-norms. Just take all semi-norms on $X$ composed with one of the projections.

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Thanks a lot for your reply. Thus if the dual space of $X$ is denoted by $X^{\ast}$, then the dual space of $X^{\infty}$ is $(X^{\ast})^{\infty}$, is it right? – CodeGolf Jan 22 '14 at 9:05
Certainly not! Why should that be true? – abx Jan 22 '14 at 9:18
Sorry I made a mistake, it is not true. – CodeGolf Jan 22 '14 at 9:22
The dual of the product $X^\infty$ is the countable direct sum of the duals. – Jochen Wengenroth Jan 28 '14 at 7:25

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