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In C.H. Edwards's Advanced Calculus of Several Variables he defines the ordinate set $\mathcal{O}_f$ of a function $f:\mathbb{R}^n\to\mathbb{R}$ as the set of points between $\mathbb{R}^n$ and the graph of $f,$ including the points of evaluation, $\mathbf{x}\in\mathbb{R}^n$ and the points in the graph $\left\{\mathbf{x},f\left(\mathbf{x}\right)\right\}\in\mathbb{R}^{n+1}$. Later on he defines a set $\hat{\mathcal{G}}=\partial\mathcal{O}_f-\mathbb{R}^n,$ where $\partial\mathcal{O}_f$ is the boundary of $\mathcal{O}_f.$ The intent seems clear. First

$$\mathbb{R}^{n+1}-\mathbb{R}^n=\mathbb{R}^n\times\left(\mathbb{R}-\left\{0\right\}\right)$$

where $\times$ means Cartesian product. Then

$$\hat{\mathcal{G}}=\left(\mathbb{R}^{n+1}-\mathbb{R}^n\right)\cap\partial\mathcal{O}_f.$$

But long ago I learned that $\mathbb{R}^n$ is the set of all real number n-tuples, and $\mathbb{R}^{n+1}$ is the set of all (n+1)-tuples, so elements of $\mathbb{R}^{n}$ are not elements of $\mathbb{R}^{n+1}$ and $\mathbb{R}^{n}$ is not a subset of $\mathbb{R}^{n+1}.$

So am I correct in concluding that $\mathbb{R}^{n+1}-\mathbb{R}^n$ is not really the relative complement of the two sets?

Steven Thomas Hatton
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  • To answer the title question, no; however, it is a common abuse of notation to write $\mathbb R^m$ as the $\mathbb R$-subspace of $\mathbb R^n$ (with $1 \leq m \leq n - 1$) given by the direct sum of $\mathbb R^m$ and $n - m$ copies of ${0}.$ Of course, direct sums of vector spaces commute, so order doesn't matter. As the below answer suggests, we can think of the $x$- and $y$-axes of $\mathbb R^2$ as the real number line $\mathbb R$ because we have that $\mathbb R \cong \mathbb R \oplus {0} \cong {0} \oplus \mathbb R,$ and the latter two are $\mathbb R$-subspaces of $\mathbb R^2.$ – Dylan C. Beck Mar 04 '21 at 19:58
  • What you'll really like is the definition of $\mathbb R^\infty$: first one writes the infinite increasing sequence $$\mathbb R \subset \mathbb R^2 \subset \mathbb R^3 \subset \mathbb R^4 \subset \cdots$$ and then one defines $$\mathbb R^\infty = \bigcup_{n=1}^\infty \mathbb R^n$$ – Lee Mosher Mar 04 '21 at 21:12

1 Answers1

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Technically you are correct, but it s customary to identify $\mathbb{R}^n$ with its natural embedding into $\mathbb{R}^{n+1}$, just as the real line is identified with the $x$-axis, (or the $y$-axis) in $\mathbb{R}^2$.

uniquesolution
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  • Edwards would have done well to elaborate on this subtlety, at least to the point of acknowledging his subterfuge. The situation evokes countless ideas and questions. Not the least of which is: so much for basis agnosticism. – Steven Thomas Hatton Mar 05 '21 at 16:46
  • I'm not going to change my acceptance, but the way Edwards defines things, I believe $\mathbb{R}^n-\mathbb{R}^m$ is a legitimate set operation. $\mathbb{R}^m$ is the subspace of $\mathbb{R}^n$ spanned by the first $m$ standard basis vectors. As a subspace of $\mathbb{R}^n$ it is the set of all vectors with the components $r^{i>m}=0$. A subspace is a subset. – Steven Thomas Hatton Mar 09 '21 at 17:56