I'm new to coalgebras and this is a question from section 2.1 of the book "Hopf Algebras" from Davied E Radford.
I tried to pick an element $u \in U$, so $\Delta(u) = u_1\otimes u_2 \in V\otimes W$ For the fact that $V$ and $W$ are subspaces we have that $V\otimes W$ is a vector space and it has a basis, so we can write $\Delta(u) = u_1\otimes u_2$ = $\sum_{}^{} a_{ij}v_i\otimes w_j$
So, i tried to use the sigma notation on $\Delta(u)$ (because i thought it could make things easier), and then i got $$\sum_{}u_1\otimes u_2 = \sum_{}^{} a_{ij}v_i\otimes w_j$$ this is the same as $\sum_{}\sum_{}(a_{ij}v_i\otimes w_j - u_1\otimes u_2)=0$
Thus $$a_{ij}v_i\otimes w_j - u_1\otimes u_2=0$$ moreover, $a_{ij}v_i\otimes w_j = u_1\otimes u_2$
I did those calculations (and i am not so sure if they are really right) and i couldn't get anywhere after that. I would be very grateful if anyone could give me an idea on how to aproach this problem.
Start fixing a base $\{ v_i \}_{i \in I}$ for $V$ and $\{ w_j \}_{j \in J}$ for $W$ and write: $$ \Delta(u) = \sum_{i,j} \alpha_{ij}v_i \otimes w_j \hspace{50px}(1)$$ By the counity axiom, it follows that: $$ u = (\epsilon \otimes id)(\Delta(u)) = \sum_{i,j} \alpha_{ij}\epsilon(v_i) \otimes w_j = \sum_{i,j} \alpha_{ij}\epsilon(v_i)w_j $$ Where in the last step we use the canonical isomorphism $k \times C \simeq C$. Since $\epsilon : C \to k$, $\alpha_{ij}\epsilon(v_i) \in k$, so it means that $u \in \text{Span}\{w_j\}$. In the same way, applying $(id \otimes \epsilon)$ to both sides of $(1)$, you can use the axioms $(id \otimes \epsilon)(\Delta(u)) = u$ and $C \otimes k \simeq C$ to get $u \in \text{Span}\{v_i\}$, so it must be $U \in V \cap W$.