I am currently reading Vector Calculus, Linear Algebra, and Differential Forms by John Hubbard and Barbara Hubbard and am having a bit of trouble reconciling notation with definition. The book is trying to express how to set up arithmetic for the real numbers and in doing so mentions a function which I can't quite seem to follow.

It doesn't seem like I can upload an image so this might be a bit messy.

If we have a function $f: D^n\to D$ where $D$ is the set of finite decimals and a function $g:R^n\to R$ where $R$ is the set of real numbers. Then the book defines a function $$g(X) = \sup_k \inf_{l\geq k}f([X]_l)$$

$X$ is an n-tuple in $R^n$. $[X]_l$ is the n-tuple with each value truncated to the l th value after the decimal. My problem comes in at the supremum and infimum. I am not sure what these are saying about the function. I know that the supremum and infimum are the least upper bound and greatest lower bound respectively, however I am not sure of their role in this function.

I am thinking it means no value of the $f()$ is larger than $k$, but then this would also mean no value of $f()$ is smaller than $l$ which is equal to (or larger than?) $k$. Which just seems weird. How would $sup$ and $inf$ apply to a function they precede?

Sorry for text mess up there. :x


We are given a function $f:\>D^n\to D$ where $D$ is the set of finite decimals. This $f$ accepts any $n$-tuple ${\bf d}=(d_1,d_2,\ldots, d_n)$ of finite decimals as input and produces a finite decimal $f({\bf d})$ as output. An example: $$f(0.21,\ {-3},\ 4.533371)=12.62\ .$$ It seems that there no further assumptions about this $f$ are made. In the following I shall assume that it is bounded.

Using this $f$ and the truncation operators $$T_l:\quad {\mathbb R}^n\to D^n,\qquad {\bf x}\to[{\bf x}]_l$$ a new function $g:\>{\mathbb R}^n\to{\mathbb R}$ is created as follows: Let an ${\bf x}\in{\mathbb R}^n$ be given. For each $l\geq0$ the number $\alpha_l:=f\bigl([{\bf x}]_l\bigr)$ is a finite decimal, hopefully with more decimal places as $l$ increases. We fix a "gauge" $k$ and consider only the $\alpha_l$ with $l\geq k$. The $\inf$ of these $\alpha_l$, i.e., the number $$\rho_k:=\inf_{l\geq k}f\bigl([{\bf x}]_l\bigr)$$ is a weak form of limit of the $f\bigl([{\bf x}]_l\bigr)$ when $l\to\infty$. But taking the $\inf$, which we can do without any further assumptions, has a serious drawback: If, e.g., the $\alpha_l$ were actually increasing with $l$ then $\rho_k=\alpha_k$, and the increased precision when $l\to\infty$ is lost. That's where the gauge $k$ comes in: The $\rho_k$ contain only information about the $\alpha_l$ with $l\geq k$; furthermore they form an increasing sequence. It follows that $$g({\bf x}):=\lim_{k\to\infty}\rho_k=\lim_{k\to\infty}\>\inf_{l\geq k}f\bigl([{\bf x}]_l\bigr) $$ is on the one hand a well defined real number, and on the other hand the $\alpha_l$ with small $l$ have no influence on the value $g({\bf x})$ anymore. (Remark: Since $(\rho_k)_{k\geq0}$ is an increasing sequence one has $\lim_{k\to\infty}\rho_k=\sup_k\rho_k$.)

Don't ask me why the Hubbards set such a thing up. I guess their motivation would come out of the context.

  • $\begingroup$ Thank you for the thorough dissection of the function as it definitely helped. My only question is where the limit involving k comes in as it seems as though a supremum k should be there. The purpose of g is to show that there is a unique function for each f that satisfies the continuity condition. $\endgroup$ – OhLawdyLawdy Jun 28 '15 at 13:45

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