Bounded neighbourhood of zero in TVS

Question

Is is true that in any topological vector space, which is $T_1$ there exists bounded neighbourhood of zero ? Is is still true if we omit $T_1$ axiom ?

Answer 1

A topological vector space in which there is a bounded neighbourhood of $0$ is called locally bounded. Whether the TVS $X$ is $T_1$ or not doesn't play much of a role, $X$ is locally bounded if and only if its associated Hausdorff TVS $X/\overline{\{0\}}$ is locally bounded. If we're only considering vector spaces over $\mathbb{R}$ or $\mathbb{C}$, it follows that all TVSs in which $\overline{\{0\}}$ has finite codimension are locally bounded, so if "$T_1$ fails strong enough", the space is locally bounded, but if it doesn't fail that strongly, it need not be.

For the important class of Hausdorff locally convex spaces (over $\mathbb{R}$ or $\mathbb{C}$) we have the equivalence

$$X \text{ is locally bounded} \iff X \text{ is normable.}$$

So there are a lot of interesting spaces that are not locally bounded.

Non-normable Fréchet spaces: the space of continuous functions on a nonempty open subset of $\mathbb{R}^d$ with the topology of compact convergence, the space of holomorphic functions on a nonempty open subset of $\mathbb{C}^d$ with the topology of compact convergence, the space of infinitely differentiable functions on a nonempty open subset of $\mathbb{R}^d$ with the topology of compact convergence of all derivatives. All these spaces have the Montel (or Heine-Borel) property, every closed and bounded subset of these spaces is compact.

Further, if $X$ is an infinite-dimensional normed space, and $X_w$ denotes the same space endowed with the weak topology, then $X_w$ is not locally bounded. Analogously, the dual of $X$ endowed with the weak$^{\ast}$ topology is not locally bounded.