Prove that $V_j$ is finite-dimensional for each $j = 1,…m$

Suppose $V_1,\ldots,V_m$ are vector spaces that $V_1 \times \cdots \times V_m$ is finite dimensional. Prove that $V_j$ is finite-dimensional for each $j = 1,\ldots, m$

So far I have considered proof by contradiction.

Suppose $V_j$ is infinite dimensional

• You've started out good so far. Here is the next point to consider: what kind of subset (note: subset, not subspace) can you find in an infinite-dimensional vector space that you cannot find in a finite-dimensional vector space? – Arthur Oct 19 '15 at 14:19

I can see two ways of doing this: One is the one I've hinted at in the comments section above, and it continues your line of thought for a proof by contradiction. However, I think that this is easier:

Given some $i \leq n$, the vector space $V_i$ is in a natural way a subspace of the product: for any vector $v\in V_i$, consider the element $(0,0,\ldots,0,v,0,\ldots,0)$ in the product.

Now, say the dimension of the product is $n$. Take any $n+1$ vectors $v_1, \ldots, v_{n+1} \in V_i$ and look at the corresponding vectors in the product. Since the product has dimension $n$ and there are $n+1$ vectors, they must be linearly dependent.

This shows that any $n+1$ vectors in $V_i$ are linearly dependent, and therefore $V_i$ has dimension at most $n$.

Hint: Let $p_1:V_1\times ...V_n\rightarrow V_1$ be the canonical projection. If $e_1,...,e_p$ generates $V_1\times...\times V_m$, $p_1(e_1),...,p_1(e_p)$ generates $V_1$.

For a fixed $V_j:1\leq j\leq n\$, choose a basis $\left \{ v^{(j)}_{i} \right \}_{1\leq i\leq n_{j}}$ and define $T:V_j\rightarrow V_1\times \cdots \times V_n$ by $v^{(j)}_{i}\mapsto (0,\cdots, v^{(j)}_{i},0\cdots,0)$.

Now use the fact that $T$ is injective, and Im $T$ is a subspace of $V_1\times \cdots \times V_n$ to conclude that $V_j$ is finite dimensional.