# question regarding proof for the dimension of the sum of subspaces

theorem: "Given finite dimensional subspaces $$U_1$$ and $$U_2$$ in a vector space V then: $$dim(U_1+U_2)=dim(U_1)+dim(U_2)-dim(U_1 \cap U_2)$$ "

The first step of the proof establishes that $$U_1 \cap U_2$$ is finite dimensional.

My question is why exactly is this important? Is it because the dimension of a non-finite dimensional space is undefined??

(I gather that this is of course important, just want to know the reason)

• The dimension of an infinite-dimensional space is defined (as an infinite cardinal number, a concept that you may not have come across yet). Apart from that, you need to tell us more about the proof (or maybe give a link or a reference) to help us help you: we can't say why a step in a proof is needed without seeing the proof. – Rob Arthan Apr 1 '19 at 20:54

Dimension is meaningful even when it is not finite. In general, the dimension of a vector space is defined to be the cardinality of its basis (one can show that any two bases have to share the same cardinality).

Cardinality essentially means the "size" of a set. For example: the cardinality of $$\{a,b,c\}$$ is 3, whereas the cardinality of $$\mathbb{N} = \{ 0,1,2,\dots \}$$ is $$\aleph_0$$ (i.e., countable infinity). The cardinality of $$\mathbb{R}$$ is $$2^\aleph_0$$ (i.e., continuum). We have (this can be made precise) that $$3<\aleph_0<2^\aleph_0$$. So that each of my example sets is strictly bigger than the previous one.

If you take polynomials $$\mathbb{R}[x] = \{ n \geq 0 \text{ and } a_nx^n +\cdots+a_1x+a_0 \;|\; a_0,\dots,a_n \in\mathbb{R}\}$$, you get an infinite dimensional vector space (over the reals). So many texts write: $$\dim(\mathbb{R}[x])=\infty$$.

If we are being more careful, we will specify the kind of infinite dimension we have. Specifically, $$\dim(\mathbb{R}[x])=\aleph_0$$. The reason for this is that $$\beta = \{1,x,x^2,\dots\}$$ forms a basis for the space of polynomials and there is an invertible map between $$\beta$$ and $$\mathbb{N}$$ (i.e., $$x^i \mapsto i$$).

Now while subtraction is ok for finite quantities, it is no longer ok for cardinalities. This is much like the fact that $$\infty-\infty$$ is an indeterminant form.

For example: $$\{1,2,3,\dots\}$$ take away $$\{3,4,\dots\}$$ leaves $$\{1,2\}$$, so one might conclude that $$\infty-\infty=2$$. But if we took $$\{1,2,3,\dots\}$$ and removed $$\{2,4,6,\dots\}$$ we would have $$\{1,3,5,\dots\}$$ so now we should have $$\infty-\infty=\infty$$.

So subtracting cardnalities is problematic. Adding them is not. One says that if $$A$$ has cardinal number $$\alpha$$ and $$B$$ has cardinal number $$\beta$$ then the disjoint union of $$A$$ and $$B$$ (put these sets together without allowing overlapping) has cardinal number $$\alpha+\beta$$.

For example: $$A=\{1,2,3\}$$ and $$B=\{1,2\}$$ then make $$B$$ distinct: $$B' = \{1',2'\}$$ put them together and get $$\{1,2,3,1',2'\}$$ thus $$2+3=5$$.

One can make sense of this for infinite stuff as well.

So why did you text prove finite dimensional first? Possibly because the statement involved a subtraction.

Actually if you say: $$\dim(U_1+U_2)+\dim(U_1\cap U_2) = \dim(U_1)+\dim(U_2)$$, then this statement holds without the finite dimensional assumption!

Proof sketch: Let $$\alpha$$ be a basis for $$U_1 \cap U_2$$. Extend this to a basis $$\beta = \alpha \cup \beta_0$$ (disjoint union) for $$U_1$$ and extend also to a basis $$\gamma = \alpha \cup \gamma_0$$ (disjoint union) for $$U_2$$.

You can show that $$\alpha \cup \beta_0 \cup \gamma_0$$ (disjoint union) is a basis for $$U_1+U_2$$. So that:

$$\dim(U_1+U_2)+\dim(U_1\cap U_2)=|\alpha \cup \beta_0 \cup \gamma_0| + |\alpha|$$ $$= |\alpha|+|\beta_0|+|\gamma_0|+|\alpha| = |\alpha \cup \beta_0| + |\alpha\cup\gamma_0| = \dim(U_1)+\dim(U_2)$$

[where $$|X|$$ denotes the cardinality of $$X$$. We used the fact that $$|X\cup Y| = |X|+|Y|$$ for disjoint sets and that cardinal addition is commutative and associative.]

This proof works whether the dimensions are finite or infinite. :)