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 ?

Thank you for your help.

  • $\begingroup$ 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. $\endgroup$ – s.harp Sep 9 '19 at 21:04
  • $\begingroup$ Thank you for you answer! You are right. Thank you for your help, I have modified the post. $\endgroup$ – Grelier Sep 9 '19 at 21:14

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 $\bigcap_{n\in\mathbb N}U_n=\{0\}$.

  • $\begingroup$ The new family of semi-norms does not need to be a subfamily of the family of original semi-norms. Thank you very much for you counterexample. That is what I needed. $\endgroup$ – Grelier Sep 10 '19 at 8:58

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