Dimension of quotient space of real connected closed intervals

Question

Let $\mathcal{C}$ be the set of all finite closed connected intervals of ℝ. Define the evaluation map \begin{eqnarray*} E : &\mathcal{C} &\longrightarrow ℝ \\ &[b,c] &\mapsto \int_b^c 1dx \end{eqnarray*} Now Let Span $\mathcal{C}$ be the (formal) vector space of all linear combinations of intervals in $\mathcal{C}$, and extend $E$ linearly to this space.* Consider the equivalence relation $X \sim \tilde{X} \iff E(X) = E(\tilde{X})$.

Then what is the dimension of Span $\mathcal{C} /\sim $?

I can't simply identify $\sim$ with a linear subspace $W$ of Span $\mathcal{C}$ and use the Dimension Formula for linear maps, because of the (uncountably) infinite dimensionalities. Can we maybe find a homomorphism with finite-dimensional image of which the kernel is exactly $\sim$? Or is the quotient really still an infite-dimensional vector space?

*i.e. $$E\left(\sum_{i=0}^na_i[b_i,c_i]\right) := \sum_{i=0}^n a_i \int_{b_i}^{c_i}1dx $$

Answer 1

I assume we're talking about the dimension as a vector space over $\Bbb R$.

We can take as a basis of $\operatorname{Span}(\mathcal C)$ the set $\{[b, c]~\vert~ b \in \Bbb R, c \gt b\}$. Then $\varphi: \operatorname{Span}(\mathcal C) \to \Bbb R$ generated by $\varphi ([b, c])=c-b$ is a linear map with kernel $\sim$, so $\dim (\operatorname{Span}(\mathcal C / \sim))=1.$

I should add that this is really nothing more than the observation that $E$ as extended is a linear map.

Answer 2

Define $\phi : \operatorname{span}\mathcal C/\sim\to\mathbb R$ by $\phi([X]_\sim):=E(X)$. This is a well-defined map and it is linear. And it is injective: if $\phi([X]_\sim) = 0$, then $E(X) = 0 = E(0)$, thus $[X]_\sim=0$.