This question is linked to Checking if the given structure is a vector space . To rephrase the question : $$ $$Let $F$ be $\Bbb R$ and let $V$ be the set $\Bbb R^+$ of all positive real numbers. Define the "sum" denoted by α $\oplus$ β of any two positive real numbers α and β, and define the "product" denoted by α $\odot$ β of any positive real number α by an arbitrary (not necessarily positive) real number β as follows: α $\oplus$ β = αβ and α $\odot$ β = $β^α$
$$ $$ In one proposition I am having a bit of problem! Before stating that, let me clarify why I am posting the same question again. In that particular question, clearly the question was not understood properly hence in the answer only that was clarified. What I am having is a bit more subtle : $$$$ $$(\alpha\oplus\beta)\odot x=\alpha \odot x\oplus\beta \odot x,\forall \alpha,\beta\in\mathbb F,x\in V$$ $$LHS=(\alpha\oplus\beta) \odot x=x^{\alpha\beta}$$ $$RHS=\alpha \odot x\oplus\beta \odot x=x^{\alpha}x^{\beta} = x^{\alpha * \beta} $$ my question is what will be tha value of * ? will it be $\oplus$ or + ? If it is $\oplus$, I can see that both L.H.S and R.H.S will be equal and the whole structure will indeed become a vector space as I have gone through all other propositions. But I am having trouble in understanding how $\oplus$ will also follow the laws of exponentiation! The natural value should be +(our good ol' even though not defined in the question), spitted out by the law of exponentiation and as a result the whole structure won't be a vector space. Can you please point out how to think about such things? $$$$P.S: Halmos says it is a vector space.
You refer to addition in $V$ with $\oplus$ and to multiplication with a scalar with $\odot$, and $F$ is $\mathbb{R}$ with just the usual addition and multiplication which we should denote by $+$ and $\cdot$.
And that is the point you are confusing here: The axiom you are interested in is \begin{equation*} (\alpha + \beta) \odot x = \alpha \odot x \oplus \beta \odot x \end{equation*} where $+$ is just the usual $+$ from the real numbers (as on the left hand side you add elements of the ground field whereas on the right hand side, you add vectors.)
Understanding this, it is straight forward to check the axiom: \begin{equation*} (\alpha + \beta) \odot x = x^{\alpha + \beta} = x^\alpha \cdot x^\beta = x^\alpha \oplus x^\beta = \alpha \odot x \oplus \beta \odot x. \end{equation*}