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Suppose that we have a Hausdorff locally convex space with its topology $\tau$ and let $P(X)$ be a separating family of $\tau$-continuous semi-norms so that $\tau$ is generated by $P(X)$. How do we prove the following:

If $p_1,\cdots,p_n\in P(X)$, then we can find $p\in P(X)$ and a constant $k\ge 1$ such that for each $i\in \{1,\cdots,n\}$ and each $x\in X$, we have $p_i(x)\le k\cdot p(x)$.

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    Are there other assumptions about $P(X)$? for example, if $X$ is the space of continuous functions on $[0,1]$ and $\rho_x(f):=|f(x)|$, it gives a separating family of semi-norms.2012-11-09

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A semi-normed space doesn't need to have the mentioned property. For example, take $C[0,1]$, the space of continuous functions with real values on the unit interval, and the semi-norms given by $\rho_x(f):=|f(x)|$. If $x_j,1\leqslant j\leqslant n$ are different elements of $[0,1]$ and $t\in [0,1]\setminus \{x_1,\dots,x_n\}$, we can find a continuous function which is $1$ at $x_1$ and $0$ at $t$.

However, if we assume the set of semi-norms filtrating, that is, for each finite collection of semi-norms $\{p_{i_1},\dots,p_{i_n}\}$, we can find an element $p_{i_{n+1}}$ such that for each $x\in X$, $\max_{1\leqslant j\leqslant n}p_{i_j}(x)\leq p_{i_{n+1}}(x)$.

For a vector space with semi-norms $\{p_i,i\in I\}$, we can add new semi-norms by the following way: for $F\subset I$ finite, define $\rho_F(x):=\max_{i\in F}p_i(x).$ This gives a family of semi-norms which gives the same topology as $\{p_i,i\in I\}$.

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    Going back to my previous comment whether $P(X)$ is filtrating, I think the answer is yes, with respect to the locally convex space whose topology is generated or determined by $P(X)$. The way I defined $P(X)$ and the one you suggested the construction of new semi-norm $p_F$, it follows that $p_F\in P(X)$.2012-11-11