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This is a follow up question to this one, I didn't feel like to have extended comments so I prefer to ask a separate question.

There's this comment specifically:

Suppose we have two families of seminorms $p_\alpha$ and $q_\beta$. If for every $\alpha$ there is $\beta$ and a constant $C_{\alpha,\beta}$ such that $p_\alpha \le C_{\alpha,\beta}q_\beta$ (and symmetrically for every $\beta$ there is $\alpha$) then they define the same topology. If the families are directed it suffices to consider only sufficiently "large" $\alpha,\beta$ because other seminorms are dominated by them. This is why the ordering helps

The thing is I don't know if this is some result from general topology or topological vector space, I don't have a reference for this. Can anyone provide a proof? or even just pointing out to a proof?

Update

According to @Fred's answer, I'll also use Rudin's convention to denote the subbasis of the topology generated by the seminorms, namely if $\left\{ p_{\alpha}\right\}$ is a family of seminorms then

$$ V(p_\alpha,n) := V(\alpha,x) = \left\{x : p_\alpha(x) < \frac{1}{n} \right\} $$

Is a subbase for topology on the vector space $X$. Suppose $\left\{p_\alpha\right\}$ and $\left\{q_\beta \right\}$ are two families of seminorms such that for each $\alpha, \beta$ we constants $K_{\alpha,\beta}$ and $C_{\alpha,\beta}$ such that

$$ K_{\alpha,\beta} q_\beta \leq p_\alpha \leq C_{\alpha,\beta}q_\beta $$

then the two topologies are equivalent.

My proof consist of showing they generate the same subbases (I'm not a topologist so apology in case of trivial mistakes). Suppose $x \in V_{\alpha,n}$ then since $K_{\alpha,\beta} q_\beta \leq p_\alpha$ we have

$$ p_\alpha(x) < \frac{1}{n} \Rightarrow q_\beta(x) < \frac{1}{n K_{\alpha,\beta}} < \frac{1}{\lfloor n K_{\alpha,\beta} \rfloor} $$

hence we have (for some $\beta$)

$$ V(\alpha,n) \subset V(\beta,\lfloor n K_{\alpha,\beta} \rfloor) $$

Likewise we have (for some $\alpha$)

$$ V(\beta,m) \subset V(\alpha,\lfloor m C_{\alpha,\beta} \rfloor) $$

I think then we can conclude from these inclusions that the two generated topologies are the same (because they generate the same subbasis essentially).

Am I wrong?

given this answer I believe I'm right

user8469759
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1 Answers1

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Let $V$ be a local convex topological vector space and let us denote the systems of seminorms by $\{p_{\alpha}: \alpha \in A\}$ and $\{q_{\beta}: \beta\in B\}$.

The topology on $V$ defined by $\{p_{\alpha}: \alpha \in A\}$ is given by the base of neighborhoods of $0$ :

$$U_{\epsilon, A'}=\{x \in V: p_{\alpha}(x)< \epsilon: \alpha \in A'\},$$

where $ \epsilon >0$ and $A'$ is a finite subset of $A$. The same holds for the topology defined by $\{q_{\beta}: \beta\in B\}$.

From $p_\alpha ≤ C_{\alpha,\beta}q_\beta$ it is now easy to see that $\{p_{\alpha}: \alpha \in A\}$ and $\{q_{\beta}: \beta\in B\}$ define the same topology on $V$.

Fred
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