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 $p_i(x)\le k\cdot p(x)$.
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 $\{p_i,i\in I\}$.