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This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ need not be connected contractible themselves, or even connected (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of sets in a trivializing cover for your bundle $E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a continuous local section (that is, a continuous map $s\colon\thinspace U_i\to E$ such that $p\circ s =\operatorname{incl}\colon\thinspace U_i\hookrightarrow B$).

Then the minimum number of sets in a trivializing cover for the vector bundle $E\to B$ equals the sectional category of the frame bundle of $F(E)\to B$.

Addendum: One can show using obstruction theory that for a $r$-connected manifold $B$, $$\operatorname{cat}(B)< \frac{\dim(B)+1}{r+1}+1.$$

So if, say, your manifold is simply-connected, you can find a trivializing cover with roughly half as many sets.

3 added 173 characters in body

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ need not be connected themselves (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of sets in a trivializing cover for your bundle $E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a continuous local section (that is, a continuous map $s\colon\thinspace U_i\to E$ such that $p\circ s =\operatorname{incl}\colon\thinspace U_i\hookrightarrow B$).

Then the minimum number of sets in a trivializing cover for the vector bundle $E\to B$ is equals the sectional category of the frame bundle of $F(E)\to B$.

2 deleted 32 characters in body

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ are need not necessarily be connected themselves (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater then than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of open sets needed to in a trivializing cover $B$ such that for your bundle $\xi$ is trivial on each, E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category$\operatorname{secat}(p)$of a fibre bundle$p\colon\thinspace E\to B$is the smallest number$k$such that$B$has a cover by open sets$U_1,\ldots , U_k$on each of which$p$admits a local section. Then the minimum number of sets in a trivializing cover for$\xi$E\to B$ is the sectional category of the associated frame bundle of $\xi$. F(E)\to B\$.

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