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Let $X$ be a countable set. Consider the space $\mathbb{R}^X$ of real-valued functions on $X$ equipped with the product topology.

Is $\mathbb{R}^X$ locally convex?

If not, is the space of real-valued bounded functions on X?

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The product topology is always locally convex if the topologies in each fibre are locally convex. The space $\prod_{x\in X} Y_x$, where each $Y_x$ is a locally convex space, is locally convex. This is due to the fact that basic neighbourhoods are given by products $\prod_{x\in X} V_x$, where finitely many satisfy $V_x\ne Y_x$, and $V_x$ is open. If each fibre is locally convex, given a point we may choose convex neighbourhoods $V_x'\subset V_x$ for the nontrivial components, and then $\prod_{x\in X}V_x'$ is a convex neighbourhood.

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It’s clearly locally convex. The basic open neighbourhoods on this product (can be any size) are convex, taking intervals at the non-trivial coordinates.

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