Topological supremum of family of linear topologies In Hans Jarchow - Locally Convex Spaces 2.4.4 (c) it says:

The topological supremum of any family of linear topologies on a fixed vector space is linear.

I couldn't find a proof in the book and neither succeeded to proof it myself, any ideas?
OK, I didn't thought about it the way Tom Cruise mentioned, the supremum as the topology generated by the union.  I guess it's pretty much straight forward now, let's say we have the linear topologies $\tau_\alpha$ on the vector space $X$ and $\tau$ is the supremum of the $\tau_\alpha$, i.e. the topology generated by sets of the form $\cap_{i&lt;n} A_{\alpha(i)}$ with $A_{\alpha(i)}\in\tau_{\alpha(i)}$.  

To see that $\tau$ is linear we have to check that it is translation invariant and contains a neighbourhood base ${\cal U}$ around $0$ of balanced, absorbing sets such that $$\forall U\in{\cal U}.\exists V\in{\cal U}.V+V\subseteq U.$$

The basic open sets are translation invariant since $\cap_{i&lt;n}A_{\alpha(i)}+x=\cap_{i&lt;n}(A_{\alpha(i)}+x)$ and one can show similarly that the collection of arbitrary unions of basic open sets, i.e. $\tau$, is translation invariant as well.

Then choose neighbourhood bases ${\cal U}_\alpha$ of $0$ in $\tau_\alpha$ with the properties mentioned above and set ${\cal U}$ as the collection of sets $\cap_{i&lt;n}U_{\alpha(i)}$ with $U_{\alpha(i)}\in{\cal U}_{\alpha(i)}$.  Obviously this is a neighbourhood base in $\tau$ around $0$ which is also balanced and absorbing and if we choose $V_{\alpha(i)}\in{\cal U}_{\alpha(i)}$ with $V_{\alpha(i)}+V_{\alpha(i)}\subseteq U_{\alpha(i)}$ then we get $$\cap_{i&lt;n} V_{\alpha(i)}+\cap_{i&lt;n} V_{\alpha(i)}\subseteq \cap_{i&lt;n} U_{\alpha(i)}$$ which completes the proof.
Jarchow does not provide a proof of Proposition 2.4.4(c) because it follows directly from Proposition 2.4.1. Indeed, if $\{\mathscr T_j\}_{j \in J}$ is an arbitrary collection of linear topologies on $E$, then the topological supremum of this family is exactly the initial (i.e. &quot;projective&quot;, &quot;weak&quot;) topology induced by the family of linear maps $\{\text{id} : E \to (E,\mathscr T_j)\}_{j\in J}$.

Added later: my answer was downvoted without any kind of explanation, so I'm guessing that the real problem lies not with the deduction of Proposition 2.4.4(c) but with the proof of Proposition 2.4.1.
The proof is given directly before the statement of Proposition 2.4.1, and it is very terse. It is therefore understandable that one may have missed or misunderstood it. Here is an expanded version of Jarchow's argument:
Proposition 2.4.1. Let $E$ be a vector space, and let $\{T_j\}_{j\in J}$ be a family of linear maps $T_j : E \to E_j$, where each $E_j$ is a tvs. Then the initial topology on $E$ for the maps $\{T_j\}_{j\in J}$ is a linear topology.
Proof (following Jarchow's argument). We have to prove that addition $S_E : E \times E \to E$ and scalar multiplication $M_E : \mathbb{K} \times E \to E$ are continuous with respect to the corresponding product topologies. Jarchow proves this simply by pointing out the identities $S_{E_j} \circ (T_j \times T_j) = T_j \circ S_E$ and $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j) = T_j \circ M_E$. The implication here is that these identities show that $T_j \circ S_E$ and $T_j \circ M_E$ are continuous for all $j\in J$, so that it follows from the characteristic property of the initial topology that $S_E$ and $M_E$ are continuous (which is what we had to show). Thus, to complete the proof, we must verify that:

$S_{E_j} \circ (T_j \times T_j)$ and $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j)$ are continuous;
$S_{E_j} \circ (T_j \times T_j) = T_j \circ S_E$ and $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j) = T_j \circ M_E$.

Claim 1. The map $T_j \times T_j : E \times E \to E_j \times E_j$, $(x,y) \mapsto (T_jx,T_jy)$ is continuous. Proof: by the universal property of the product topology, it suffices to show that $(x,y) \mapsto T_jx$ and $(x,y) \mapsto T_jy$ are continuous. But this is trivial, since these maps are given by the compositions $T_j \circ \pi_x$ and $T_j \circ \pi_y$ of continuous functions.
Claim 2. The map $\text{id}_{\mathbb{K}} \times T_j : \mathbb{K} \times E \to \mathbb{K} \times E_j$, $(\lambda,x) \mapsto (\lambda,T_jx)$ is continuous. Proof: analogous.
Since $S_{E_j}$ and $M_{E_j}$ are continuous (by assumption), it follows that $S_{E_j} \circ (T_j \times T_j)$ and $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j)$ are continuous. It remains to check the claimed identities.
Claim 3. One has $S_{E_j} \circ (T_j \times T_j) = T_j \circ S_E$. Proof: both compositions are maps $E \times E \to E_j$, so the domain and codomain match. Furthermore, for all $(x,y) \in E \times E$ we have
\begin{align*}
 (S_{E_j} \circ (T_j \times T_j))(x,y) &amp;= S_{E_j}((T_j \times T_j)(x,y)) = S_{E_j}(T_jx,T_jy) = T_jx + T_jy\\
&amp;= T_j(x+y) = T_j(S_E(x,y)) = (T_j \circ S_E)(x,y).
\end{align*}
Claim 4. One has $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j) = T_j \circ M_E$. Proof: both compositions are maps $\mathbb{K} \times E \to E_j$, so the domain and codomain match. Furthermore, for all $(\lambda,x) \in \mathbb{K} \times E$ we have
\begin{align*}
 (M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j))(\lambda,x) &amp;= M_{E_j}((\text{id}_{\mathbb{K}} \times T_j)(\lambda,x)) = M_{E_j}(\lambda,T_jx) = \lambda T_j x\\
&amp;= T_j(\lambda x) = T_j(M_E(\lambda,x)) = (T_j \circ M_E)(\lambda,x).\tag*{$\Box$}
\end{align*}