A $\Bbb K$ vector bundle over a space $B$ is a space $E$ and a continuous map $p:E\to B$ so that $p^{-1}(b)$ is a topological $\Bbb K$ vector space for any $b\in B$.
One always includes local triviality in the definition, that is one demands that for any $b\in B$ there is a neighbourhood $U$ of $B$ together with homeomorphism $f:U\times \Bbb R^n \to p^{-1}(U)$ so that $p\circ f$ is the same as the projection onto $U$.
If one does not include this in the definition, can one construct simple examples of vector bundles that are not locally trivial?
Best Answer
$\newcommand{\Reals}{\mathbf{R}}$To summarize (and slightly amplify) the comments: Define a vector party (thinking of a disorganized collection, with or without local triviality) over a space $B$ to be a space $E$ together with a continuous map $p:E \to B$ such that $p^{-1}(b)$ is a vector space (over some fixed field) for each $b$ in $B$.
A vector party is far from a vector bundle, merely a collection of vector spaces parametrized by points of $B$. Fix a topological vector space $V$ of dimension at least $2$ (but not necessarily finite).
A vector party can fail to be a vector bundle for multiple reasons, including:
The total space $E$ has "finer topology than it should".
Let $E = B \times V$ be the Cartesian product with the discrete topology on $B$, and $p:E \to B$ projection to the first factor. If $O$ is an open set in $V$, then for each $b$ in $B$, the set $\{b\} \times O$ is open in $E$; if the base space $B$ is not discrete, $p:E \to B$ is not locally trivial. (In general, $E$ has no continuous, non-constant local section.)
The fibres "vary discontinuously".
Let $k$ be an integer, $0 < k < \dim V$, let $\sigma:B \to G_{k}(V)$ be an arbitrary mapping from $B$ to the Grassmannian of $k$-planes in $V$ (i.e., "a choice of $k$-dimensional subspace of $V$ for each $b$ in $B$"), let $E$ be the collection $$ E = \bigcup_{b \in B} \{b\} \times \sigma(b) \subset B \times V $$ equipped with the subspace topology, and let $p:E \to B$ be (the restriction of) projection to the first factor.
If $\sigma$ is discontinuous (e.g., map $\Reals$ to the space of lines through the origin in $\Reals^{2}$ by taking $\sigma(b)$ to be the $x$-axis if $b$ is rational, the $y$-axis if $b$ is irrational, as in the comments), then the subspace topology on $E$ differs from the "local product" topology of a vector bundle. (In general, $E$ has no continuous, non-trivial local section.)
The fibres "vary continuously", but "the dimension jumps".
If $E$ and $E'$ are (smooth, say) vector bundles over $B$, and if $f:E \to E'$ is a (smooth) bundle morphism, the "bundle of kernels" (whose fibre at $b$ is the kernel of $f_{b}:E_{b} \to E'_{b}$) and the "bundle of images" are vector parties, but generally not vector bundles.
Think, for example, of the helicoid $E \subset \Reals \times \Reals^{2}$ parametrized by $$ (b, v) \mapsto \bigl(b, (v\cos b, v\sin b)\bigr), $$ the product $E' = \Reals \times (\{0\} \times \Reals)$, a.k.a., the $(x, z)$-plane, and the bundle morphism $f:E \to E'$ given by projection to the $(x, z)$-plane, i.e., $$ \bigl(b, (v\cos b, v\sin b)\bigr) \mapsto \bigl(b, (0, v\sin b)\bigr). $$ The "party of kernels" is the bundle analog of the skyscraper sheaf supported at integer multiples of $\pi$, while the "party of images" results from collapsing these fibres of $E'$ to points.