Proof of The Real Vector Space of Real-Valued Functions on a Set

Each result cited is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement above.

Claim 1. Two maps $X\to\mathbb{R}$ are equal exactly when they take the same value at every point of $X$. Hence each of the conditions 1 to 8 of Vector Space over a Field, for the set $\mathrm{Map}(X,\mathbb{R})$ over the field $\mathbb{R}$ of Field, is equivalent to the corresponding identity between real numbers holding at every point. Let $f,g,h\in\mathrm{Map}(X,\mathbb{R})$, let $\lambda,\mu\in\mathbb{R}$ and let $x\in X$. Writing out the two operations, and using in each case the field axiom of $\mathbb{R}$ of the same name applied to the real numbers $f(x),g(x),h(x),\lambda,\mu$:

This gives conditions 1, 2, 3, 5, 6, 7 and 8, condition 3 with the element $0_{X}$. For condition 4, let $n_{f}$ be the map $x\mapsto -f(x)$, where $-f(x)$ is the additive inverse of $f(x)$ in $\mathbb{R}$; then $(f+n_{f})(x)=f(x)+(-f(x))=0=0_{X}(x)$ for every $x$, so $f+n_{f}=0_{X}$. Hence $\mathrm{Map}(X,\mathbb{R})$, with the pointwise operations, is a vector space over $\mathbb{R}$.

By claim 1 of Elementary Identities in a Vector Space a vector space has exactly one zero vector, and $0_{X}$ is one by the third display, so the zero vector is $0_{X}$. By claim 2 of that lemma each $f$ has exactly one additive inverse, and $n_{f}$ is one, so $-f=n_{f}$ is the map $x\mapsto -f(x)$. Therefore $f-g=f+(-g)$ is the map sending $x$ to $f(x)+(-g(x))=f(x)-g(x)$.

Claim 2. The three hypotheses on $W$ are precisely conditions 1, 2 and 3 of Linear Subspace for the subset $W$ of the vector space $\mathrm{Map}(X,\mathbb{R})$ of claim 1, so $W$ is a linear subspace of it.

Because $W$ is closed under the pointwise sum and the pointwise scalar multiple, the restrictions of these two operations to $W$ take their values in $W$ and are therefore operations on $W$ of the kind required by Vector Space over a Field. Conditions 1, 2, 5, 6, 7 and 8 are identities between elements of $\mathrm{Map}(X,\mathbb{R})$ built from elements of $W$ using these operations, hence hold in $W$ because they hold in $\mathrm{Map}(X,\mathbb{R})$ by claim 1. Condition 3 holds with the element $0_{X}$, which lies in $W$ by hypothesis. For condition 4, let $f\in W$; then $(-1)f\in W$ by closure under scalar multiples, and $(-1)f=-f$ by claim 5 of Elementary Identities in a Vector Space, so $f+(-1)f=0_{X}$. Hence $W$ with the restricted operations is a vector space over $\mathbb{R}$, and its zero vector is $0_{X}$ by claim 1 of Elementary Identities in a Vector Space.