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Is there a (valid) formula for $\dim(U + V + W)$? I know from MO that $$\begin{align*} \dim(U + V + W) &= \dim(U) + \dim(V) + \dim(W)\\ &\qquad\mathop{-} \dim(U \cap V) - \dim(U \cap W) - \dim(V \cap W)\\&\qquad \mathop{+} \dim(U \cap V \cap W) \end{align*}$$ is wrong.

Can we relate $\dim(U + V + W)$ with the cardinality of some of their quotient spaces? (sorry if this is a dummy question but I'm not any familiar with quotient spaces).

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    What you describe as "the formula from the cardinality of a finite union of sets" is known as the "Inclusion-Exclusion formula", and that is **precisely** what leads to the formula that you note is incorrect.2011-04-25
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    I don't understand your question. The wrong formula **is** derived from Inclusion-Exclusion. What is it you are "wondering"? You add each dimension, then you try to "take away" the dimensions of the 2-way intersections (intuitively, what you've counted twice), then add back in the triple intersection (intuitively, what you've taken away three times after adding twice). But it doesn't work for the reasons discussed in MO.2011-04-25
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    What do you mean by "formula"? Presumably you want $dim(U+V+W)$ on the left side, but what do you allow on the right side?2011-04-25
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    I want to calculate the dimension of $U + V + W$2011-04-25
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    Presumably, Vicfred wants to write $\dim(U+V+W)$ as a linear combination of $\dim(U)$, $\dim(V)$, $\dim(W)$, $\dim(U\cap W)$, $\dim(U\cap V)$, $\dim(V \cap W)$, and $\dim(U \cap V \cap W)$.2011-04-25

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I don't think you can do better than this: $$\begin{align*} \dim(U +V + W) &= \dim((U +V) + W) \\ &= \dim(U +V) + \dim W - \dim((U+V)\cap W) \\ &= \dim U + \dim V - \dim (U \cap V) + \dim W - \dim((U+V)\cap W) \end{align*}$$ Now you're stuck with $\dim((U+V)\cap W) $, for which there does not seem to be a simple formula.

(BTW, see also https://mathoverflow.net/questions/17740/is-there-a-version-of-inclusion-exclusion-for-vector-spaces but you probably know about this already.)