I have just showed that $A+B$ is indeed a subspace by showing it is non empty and closed under multiplication and addition, however I am not sure how to do these types of problems with $\mathbf{dim}$.
Problem: Dimensions
$A+B=\left \{ a+b : a\in A, b\in B \right \}$ and $C$ is a vectorspace and $A \subset C, B \subset C$.
Let $a_1,..,a_n$ be elements of $A$ and let $b_1,..,b_k$ be elements of $B$ be two families of vectors in $C$. Let $A=\operatorname{Span}(a_1,..,a_n)$ and $b=\operatorname{Span}(b_1,..,b_k)$
Now, Show that $\mathbf{dim}(A+B)\leq n+k$ and show that $\mathbf{dim}(A+B)\leq \mathbf{dim}(C)$
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I am new to the subject and I could use some help since I am really struggling with both span and dim. I get the definitions but no idea on how to approach such a problem.
Let $C$ be vector space over field $K$. You have proved that $A+B$ is subspace of $C$, then it is itself linear space over $K$. $$A=\operatorname{Span}(a_1,...,a_n)=\bigg\{\sum\limits_{t=1}^{n}\alpha_ta_t\bigg|~\alpha_t \in K\bigg\}.$$ $$B=\operatorname{Span}(b_1,...,b_k)=\biggl\{\sum\limits_{t=1}^{k}\beta_tb_t\;\biggm|~\beta_t \in K\biggr\}.$$ $$A+B=\operatorname{Span}(a_1,...,a_n)+\operatorname{Span}(b_1,...,b_k)=\biggl\{\sum\limits_{t=1}^{n}\alpha_ta_t+\sum\limits_{i=1}^{k}\beta_ib_i\;\biggm|~\alpha_t,\beta_i \in K\biggr\}=\biggl\{\sum\limits_{t=1}^{n+k}\gamma_tc_t\:\biggm|~\gamma_t \in K\biggr\},$$ where $c_t=a_t$ if $t \leq n$ and $c_t=b_{t-n}$, if $n+1 \leq t \leq n+k$. So basis of $A+B$ contains no more than $n+k$ vectors. Therefore $\dim(A+B) \leq n+k$.
2)If $\dim(C)=\infty$, then inequlity $\dim(A+B) \leq \dim(C)$ is obious. Let $\dim(C) < \infty$ and $\{e_1,...,e_s\}$ be basis of $A+B$. Hence, $\{e_1,...,e_s\}$ is linearly independent in $C$ over $K$. Basis of $C$ is maximal linearly independent system of vectors from $C$ and $\{e_1,...,e_s\}$ is some linearly independent system. Therefore $\dim(A+B)=s \leq \dim(C)$.