# Semi-norms and Hausdorff compact convex set in locally convex space

Let $$(X,\tau)$$ be an infinite dimensionnal locally convex space (no necessarily Haussdorf) and let $$K$$ be a Haussdorf compact convex subset of $$X$$. Suppose that the topology of $$X$$ is generated by a family of semi-norms $$(p_{\gamma})_{\gamma}$$ of cardinal $$\Gamma$$. Does it exist a family of semi-norms $$(p_\theta)_{\theta}$$ of cardinal $$\Theta$$ defined on $$X$$ such that:

i) $$\Theta<\Gamma$$

ii) the topology that induces on $$K$$ is smaller (or equal) than the topology induced by $$\tau$$ in $$K$$

iii) $$K$$ is still Haussdorf with this topology ?

• If $\Gamma=1$ (for example if the topology is given by a norm) then the statement is false. Also unless $\Gamma$ is the first ordinal of a given cardinality you can re-index the semi-norms in a lesser ordinal of the same cardinality, so the question is really one about cardinalities I'd guess. – s.harp Sep 9 '19 at 21:04
If I understand your question properly the answer is negative. Let me first mention that a compact Hausdorff space does not admit a strictly coarser Hausdorff topology so that ii) & iii) can be replaced by the condition that the new topology (here it is not clear to me whether you want some family of different seminorms or a subfamily of the given one) coincides on $$K$$ with the one induced by $$\{p_\gamma: \gamma\in \Gamma\}$$.
Consider the following example: Let $$I$$ be an index set of cardinality $$\aleph_1$$ and endow $$X=\mathbb R^I$$ with the product topology $$\tau$$. Then $$K=[-1,1]^I$$ is compact by Tychonov's theorem. If $$\{q_\theta:\theta\in\Theta\}$$ is a family of seminorms of strictly smaller (hence countable) cardinality then the $$0$$-neighbourhood filter of the induced topology on $$K$$ has a countable basis. Since this topology should coincide on $$K$$ with $$\tau$$ you would get a countable basis $$\{U_n:n\in\mathbb N\}$$ of $$0$$-neighbourhoods in $$K$$ for the product topology, in particular, $$\bigcap_{n\in\mathbb N}U_n=\{0\}$$. But every $$U_n$$ gives only restrictions for finitly many coordinates, i.e., there are finite sets $$I_n$$ such that $$U_n\supseteq \{(x_i)_{i\in I}\in K: x_i=0 \text{ for all }i\in I_n\}$$. Since $$J=\bigcup_{n\in\mathbb N} I_n$$ is still countable you find $$k\in I\setminus J$$ and then $$x=\delta_k$$ (i.e., $$x_k=1$$ and $$x_i=0$$ for $$i\neq k$$) defines a non-zero element of $$K$$ which belongs to $$\bigcap_{n\in\mathbb N}U_n=\{0\}$$.