About properties of inductive limit of Fréchet spaces I want to prove the following proposition about inductive limits of Fréchet spaces, from the book Grubb, Distributions and Operators:

Let $X=\bigcup_{j \in \mathbb{N}} X_{j}$ be an inductive limit of Fréchet spaces.
A set $E$ in $X$ is bounded if and only if there exists a $j_{0}$ so that $E$ lies in $X_{j_{0}}$ and is bounded there.

We use the following definition of Fréchet spaces:

A topological vector space is called a Fréchet space when $X$ is metrizable with a translation invariant metric, is complete, and is locally convex.

And the definition of boundedness:

A set $E$ in a topological vector space is said to be bounded if for every open neighborhood $U$ of the origin, there is a $t>0$ such that $E\subset tU$, where $tU = \{tx: x\in U\}$.

The book does not give a proof but says that we can consult Rudin's Functional Analysis book Chapter 6 for an "essential" proof but I cannot see what is related to this.
Any help is appreciated.
 A: Apparantly, you mean a strict LF-space, i.e., $X_n$ is a topological subspace of $X_{n+1}$ (this was the early terminology of Dieudonné and Schwartz, later Grothendieck only required continuous inclusions $X_n\hookrightarrow X_{n+1}$, this has the big advantage that the class of LF-spaces becomes stable with respect to quotients, but the regularity property for bounded sets is no longer true). In Rudin's book, only the mother of all LF-spaces $\mathscr D(\Omega)$ is considered and the proof given there only partly generalizes to the abstract setting.
Before proving the behaviour of the bounded sets, one has to clarify the topology of the inductive limit $X_\infty=\bigcup_{n\in\mathbb N} X_n$: It is the finest locally convex topology on the union so that all inclusions $X_n\hookrightarrow X_\infty$ become continuous. This is obviously generated by the set of all those seminorms $p$ on $X_\infty$ whose restrictions to all $X_n$ are continuous (for the Fréchet topology of $X_n$, more computationally, this means, $p|_{X_n}\le cp_n$ for some constant $c$ where $p_n$ is one of the seminorms given on $X_n$). Since continuity of a linear map $T:X_\infty \to Y$ for a locally convex space is equivalent to the continuity of the
seminorms $q\circ T$ for every continuous seminorm $q$ on $Y$, this implies the universal property of inductive limits: $T$ is continuous if and only if all restrictions $T|_{X_n}$ are continuous.
Using that seminorms are determined by their unit balls via the Minkowski functional (or gauge) we have that an absolutely convex set $U$ is a $0$-neighbourhood in $X_\infty$ if and only if $U\cap X_n$ is a $0$-neighbourhood in $X_n$ for all $n$.
So far, strictness was not used. Let us show that in the strict case, $X_n$ is even a toplogical subspace of $X_\infty$, for every $n$.
Indeed, given an absolutely convex $0$-neighbourhood $U_n$ in $X_n$, there is another one $V$ in $X_{n+1}$ with $V\cap X_n\subseteq U_n$, and the absolutely convex hull $U_{n+1}$ of
$U_n\cup V$ then satisfies $U_{n+1}\cap X_n=U_n$. Continuing in this way, we find absolutely convex $0$-neighbourhoods $U_m$ in $X_m$ with $U_m\cap X_{m-1}=U_{m-1}$ for all $m>n$, and then the union $U=\bigcup_{m\ge n} U_m$ is an absolutely convex (because the union is increasing) $0$-neightbourhood in $X_\infty$ (its intersection with $X_m$ contains $U_m$ for $m\ge n$ and $X_m\cap U_n$ for $m<n$) such that $U\cap X_n=U_n$.
Let now $B\subseteq X_\infty$ be bounded and assume that $B$ is not contained in any $X_n$. Just for notational convenience, we may then assume that there are $b_n\in B\cap X_{n}\setminus X_{n-1}$
(where $X_0=\{0\}$). By Hahn-Banach, there is $f_1\in X_1'$ with $f(b_1)=1$. Since $X_1$ is closed in $X_2$, again Hahn-Banach yields $f_2\in X_2'$ with $f_2(b_2)=2$ and $f_2|_{X_1}=f_1$. Continuing in this way yields $f_n\in X_n'$ with $f_{n}|_{X_{n-1}}=f_{n-1}$ and $f_n(b_n)=n$. The universal property implies that $f:X_\infty\to \mathbb K$ defined by
$f(x)=f_n(x)$ if $x\in X_n$ is well-defined and an element of $X_\infty'$. But this implies that $f(B)$ is bounded in $\mathbb K$ contradicting $f(b_n)=n$. Hence $B$ is contained in some $X_n$ and thus also bounded there because $X_n$ is a topological subspace of $X_\infty$.
You certainly realize that we did not even need that $X_n$ are Fréchet spaces. This is used for some deeper results about  LF-spaces -- but this answer is already far too long.
