Given an ordinary set $B$ and a set of indexes $I$, what is meant by $B_I$? Doing a problem set, in the context of indexed sets, I'm asked to "prove that for any sets $I$ and $B$", that $$\bigoplus B_I = B \times I$$
The $\bigoplus$ symbol is defined like so:
For an $I$-indexed set $A = \{A_i\}_{i \in I}$, we have
$$\bigoplus A = \bigoplus_{i\in I}A_i = \left\{(a,i) \in \left(\bigcup A\right) \times I \mid a \in A_i \right\}$$
Can anyone make a guess as to what is meant by $\bigoplus B_I$ given the above? I've looked through course notes but I cannot seem to find a definition for this notation....
My initial guess was that $B_I$ simply denoted that $B$ was an $I$-indexed set, but if this were the case I don't think that $B \times I$, the cross-product of an $I$-indexed set with an "ordinary" set of indexes, is defined.
edit: Doh! After seeing dtldarek's answer below I managed to find the definition in the notes:

(Constant I-indexed sets). If $I \neq \varnothing$ then, for any set $X$, the constant $I$-indexed set denoted $X_I$ is, by deﬁnition, the surjective function $$X_I \colon I \twoheadrightarrow \{X\} \colon i \mapsto X$$ 

 A: Suppose that you want to form the union of a bunch of pairwise disjoint copies of a set $B$. You can’t just choose an index set $I$ of the right cardinality, let $B_i=B$ for each $i\in I$, and form $$\bigcup_IB_I=\bigcup_{i\in I}B_i\;.\tag{1}$$ To see why, observe that $x\in\bigcup_{i\in I}B_i$ if and only if there is an $i\in I$ such that $x\in B_i$, which is the case if and only if $x\in B$. Thus, $x\in\bigcup_{i\in I}B_i$ if and only if $x\in B$, and $\bigcup_{i\in I}B_i=B$. This union just gives you $B$ back; it doesn’t give you the union of a bunch of separate copies of $B$.
The slick and easy way to make separate copies of a set is to take the Cartesian product of the set with a suitable index set. Here we use the product $B\times I$, whose elements are ordered pairs $\langle b,i\rangle$ with $b\in B$ and $i\in I$. For each $i\in I$ we gather up all the pairs with that second component and let $$B_i=\{\langle b,i\rangle\in B\times I:b\in B\}\;,$$ or simply $B_i=B\times\{i\}$. Each $B_i$ is easily thought of as a copy of $B$: we’ve just replaced each $b\in B$ with a ‘labelled $b$’, $\langle b,i\rangle$, whose label identifies which copy of $B$ contains this copy of $b$. What happens when we form the union of these $B_i$’s? 
We have $x\in\bigcup_{i\in I}B_i$ if and only if there is an $i\in I$ such that $x\in B_i$, which is the case if and only if there are a $b\in B$ and an $i\in I$ such that $x=\langle b,i\rangle$, and it follows that 
$$\bigcup_{i\in I}B_i=B\times I\;.\tag{2}$$
This union really can be thought of as the union of $|I|$ pairwise disjoint copies of $B$.
In this context the notation $\bigoplus B_I$ is an abbreviation for this process. It denotes the disjoint union of $|I|$-many copies of $B$ and implies the ‘disjointing’ procedure that I described above. Thus, it’s equal to the set in $(2)$, not the one in $(1)$.
