Wikipedia:Reference desk/Archives/Mathematics/2011 December 11

= December 11 =

Summa cum laude or magna cum laude?
norbert wiener got Tufts Univ. B.S. in 1908; also got harvard univ. phD in 1912: WAS either degree summa cum laude or magna cum laude? — Preceding unsigned comment added by 141.225.194.27 (talk) 01:03, 11 December 2011 (UTC)
 * The Tufts catalog says that his BA was cum laude, and so do other sources. As far as I know there are no such distinctions for a PhD. Looie496 (talk) 16:47, 11 December 2011 (UTC)
 * It depends on the country, university, and time frame. In Germany, PhDs are often graded "Summa cum laude", "Magna cum laude", "Cum laude", and "Rite". "Magna" is more or less the default for a PhD. The only cases of "rite" I've encountered are PhDs of politicians, usually done with the minimal effort necessary to obtain the somewhat prestigious title. Some universities have adopted more modern grades. --Stephan Schulz (talk) 16:30, 13 December 2011 (UTC)

Please help me understand my lecturer's notes - metric spaces, norms and topologies
Hello all, I'm teaching myself a course on functional analysis but having trouble understanding the notes. (The course was lectured over the last 3 months but I was taking another course which clashed so am teaching myself now.) I was hoping I could write out a section of the content and you might be able to help me figure out how these conclusions are drawn since I can't; the content from the notes is italicized below.

'' 'Metrizable' means the topology is given by a metric. We seldom use this directly; instead, we use the fact that there is a decreasing base $$(N_j)_{j=1}^{\infty}$$ of closed, absolutely convex neighbourhoods of 0. If $$p_j$$ is the gauge (Minkowski functional) of $$N_j $$ then the topology is determined by the increasing sequence $$(p_j)_{j=1}^{\infty}$$ of semi-norms. We could then take for example the metric $$\rho(x,y) = \sum_{j=1}^{\infty} \frac{p_j(x-y)}{2^j(p_j(x-y)+1)}$$. Note $$x_n \to x $$ as $$n \to \infty $$ iff $$p_j(x_n-x) \to 0 $$ as $$n \to \infty $$, for each $$j \in \mathbb{N} $$. Here are 2 easy examples: first, let $$ \omega$$ be the space of all complex sequences. Set $$p_j(x) = j \sum_{i=1}^j |x_i| $$; then the topology on $$\omega $$ is simply the product topology. Secondly, let $$C(\mathbb{R}^d)$$ be the space of all continuous functions on $$\mathbb{R}^d $$. Set $$p_j(f) = j sup_{|x| \leq j} |f(x)| $$. The resulting topology is the topology of local uniform convergence.''

So, first of all how do we conclude that the topology is simply the product/local uniform convergence topology? Are we defining seminorms $$p_j$$ (with the factor of $$j$$ at the start to ensure they're increasing), and then saying 'these must be the Minkowski functionals for some decreasing base of neighbourhoods of 0', and then using the fact that convergence under the suggested metric occurs only if convergence occurs for each $$p_j $$? For one thing, while it's obvious to me that given some set we can induce a Minkowski functional by definition, it's not obvious to me that conversely by defining a seminorm $$p_j$$, we induce some neighbourhood for which it is the Minkowski functional. Despite the fact these are meant to be 'easy' examples, I simply can't see why this collection of seminorms induces precisely the topology stated, in either case. Could anyone explain it to me? I would of course ask the lecturer but since I wasn't able to attend the course which is now over it may be difficult to clear up my confusion. Thank you very much in advance. 82.27.232.62 (talk) 02:41, 11 December 2011 (UTC)


 * Any seminorm p is the Minkowski functional for the absolutely convex set $$\{x|p(x) \leq 1\}$$ (the text also says "neighborhoods" of 0 but that seems inconsistent with the examples) . As a result, choosing an increasing sequence of seminorms is the same as choosing a decreasing sequence of absolutely convex sets and taking their Minkowski functionals.
 * You should convince yourself of the statement: "Note $$x_n \to x $$ as $$n \to \infty $$ iff $$p_j(x_n-x) \to 0 $$ as $$n \to \infty $$, for each $$j \in \mathbb{N} $$." For the first example we need to show that $$p_j(y_n) \to 0$$ as n goes to infinity for all j iff $$y_n \to 0$$ in the product topology.  Convergence in the product topology just means that the sequence in each component converges (i.e. each $$\{y_{nj}\}_{n=1}^\infty$$ converges where $$y_n = (y_{n1},y_{n2},\ldots,)$$).  If $$y_{nj}$$ doesn't go to zero for some j, then $$p_j(y_n)$$ clearly doesn't go to zero.  If $$y_{nj} \to 0$$ for all j then each p_j(y_n) converge to 0 because p_j only counts a finite number of the component sequences.  The second example works pretty much the same way. Rckrone (talk) 05:10, 12 December 2011 (UTC)


 * Ok, that all makes sense: it was silly of me not to spot the seminorm/set correspondence works both ways, thankyou. Could you clarify what you meant about 'neighborhoods' being inconsistent, sorry? I'm happy with the equivalence of convergence under the given metric iff we get convergence for all the seminorms; so I guess my final question is about the correspondence between the limits and the topology: if we know (as in this case) that convergence in the induced topology from the seminorms is the same as convergence in a different topology (such as the product topology here), why does that necessarily mean the 2 topologies themselves are the same? (How do we know there isn't some weird collection of open sets, not the same as the product topology, which gives us convergence in the seminorms iff we have convergence in the product topology? I appreciate that something being a limit of a sequence means every open set eventually contains all members of the sequence, but I can't see how it immediately follows that because we have identical limit points we have identical topologies.) Thanks again for all the help! 82.27.232.62 (talk) 06:16, 12 December 2011 (UTC)


 * In general, the limits of nets fully determine the topology of a space. If the topology is first-countable then the limits of sequences fully determine the topology, and all the topologies being dealt with here are first-countable (in particular metric topologies are first-countable).  You can show for instance that a set U is open if and only if every sequence that converges to a point in U is eventually contained in U.
 * I was wrong about the neighborhood thing. I was confused about the sets you get from the seminorms. Rckrone (talk) 07:59, 12 December 2011 (UTC)

Lone dot used in norm symbol
Is there anywhere in Wikipedia where the meaning of the lone dot in $$\|\cdot\|_V$$ is explained? The specific example that motivated this question is at Calculus_of_variations,
 * "A functional $$J(y)$$ defined on some appropriate space of functions $$V$$ with norm $$\|\cdot\|_V$$ is said to..."

Thanks. --Bob K31416 (talk) 16:10, 11 December 2011 (UTC)
 * The dot is just a placeholder for the input. Replace the dot with an element of V to denote the norm applied to that element. — Fly by Night  ( talk )  17:11, 11 December 2011 (UTC)
 * Is there an appropriate place in Wikipedia where that explanation can be placed? --Bob K31416 (talk) 18:45, 11 December 2011 (UTC)
 * Yes, it should be placed somewhere, and the right place is no doubt Function_(mathematics). I added few experimental lines at the end of that section. --pm a 19:55, 11 December 2011 (UTC)
 * Should it also be added to the meanings of · in List of mathematical symbols? Not immediately obvious how best to fit it into the table format though. Qwfp (talk) 20:19, 11 December 2011 (UTC)
 * There is no formal consensus. For example, other possibilities could be $$\|-\|_V$$ or $$\|*\|_V$$. — Fly by Night  ( talk )  22:41, 11 December 2011 (UTC)
 * In some texts I've read, simply a space is left to denote functions like this, i.e. abs = | |. That seems quite uncontentious. — Anonymous Dissident  Talk 10:14, 12 December 2011 (UTC)
 * Re consensus — Please note that it's not a matter of whether or not the dot (or anything else) should be used, which would require consensus. It's only a matter of explaining what the dot means, which doesn't require consensus since no one seems to object to Fly by Night's explanation that the dot is a placeholder. --Bob K31416 (talk) 15:37, 12 December 2011 (UTC)
 * Using the dot as a place holder does not require consensus; someone may use any notation they choose (provided they explain it). There seemed to be a suggestion of adding $$\|\cdot\|_V$$ to a notation article. That is fine, but then we would need to add $$\|*\|_V$$, $$\|-\|_V$$, $$\| \ \|_V$$, and many others. The right place for such discussions is Wikipedia_talk:WikiProject_Mathematics. — Fly by Night  ( talk )  21:38, 12 December 2011 (UTC)
 * I've never actually seen these "alternative notations" in print.  Sławomir Biały  (talk) 21:42, 12 December 2011 (UTC)
 * I've seen a dot used as a placeholder in loads of places, but I've never seen an asterisk or a dash. I've seen a space occasionally. --Tango (talk) 23:58, 12 December 2011 (UTC)

The above includes the possible addition and the two entries already there for dot. I think it might be improved, so I'm open to suggestions. --Bob K31416 (talk) 00:19, 13 December 2011 (UTC)
 * It's a lambda expression, not an element of a set. This distinguishes it syntactically from a variable.  I'm not sure how this should best be conveyed.  Sławomir Biały  (talk) 00:29, 13 December 2011 (UTC)
 * My impression is that "lambda expression" is somewhat synonymous with "placeholder", rather than the objects that can replace the placeholder or lambda expression? --Bob K31416 (talk) 01:48, 13 December 2011 (UTC)
 * Placeholder seems not quite right, since a variable x in the expression $$\|x\|$$ is just as much a placeholder, but has a different syntactic sense than a dot. The dot means the function defined by the operation, if you like.   Sławomir Biały  (talk) 14:23, 13 December 2011 (UTC)
 * I think I see the point you are trying to make with your last sentence, "The dot means the function defined by the operation, if you like." For example when one has f(x), the function itself can be referred to as f. However, when one has $$\|x\|$$, it's not as simple to refer to the norm function itself. Hence, just as one can refer to the function part f of the expression  f(x), so can one refer to the norm function part  $$ \|\cdot\| \ $$  of the expression $$ \|x\| $$.
 * Regarding your comment "Placeholder seems not quite right, since a variable x in the expression $$\|x\|$$ is just as much a placeholder, but has a different syntactic sense than a dot." — I wouldn't say that the x is just as much a placeholder as the dot, because x is an element of a set whereas the dot isn't. The expression $$\|x\|$$ has a value, whereas $$ \|\cdot\| \ $$ does not have a value. The dot is a placeholder for an element such as x but the dot is not itself an element. The dot has no more status than a blank space. I think the dot is used instead of a blank space because the blank space might be confusing, e.g. a blank might be mistaken for a typo. --Bob K31416 (talk) 04:15, 14 December 2011 (UTC)
 * No, x need not refer to a specific element of a set. Consider the notation $$y=f(x)$$ for instance.  This relates variables (indeterminates) x and y, not elements of sets.   Sławomir Biały  (talk) 12:20, 14 December 2011 (UTC)
 * The dot and x are different. Although x is a variable, not a constant, and doesn't refer to a specific element, it does represent an element of a set. (For example, "Let x be an element of the set X." Another example, "Let x be a real number.") The dot does not represent an element of a set anymore than a blank space represents an element of a set. --Bob K31416 (talk) 12:32, 14 December 2011 (UTC)
 * On second thought regarding what the dot is a placeholder for, perhaps it would be more accurate to characterize the dot as a placeholder for an argument of a function. --Bob K31416 (talk) 11:44, 14 December 2011 (UTC)
 * Alternatively, PMajer (pm a ) recently described the dot with a sentence at Function_(mathematics), "To define a function, sometimes a dot notation is used in order to emphasize the functional nature of an expression without assigning a special symbol to the variable." --Bob K31416 (talk) 12:24, 14 December 2011 (UTC)
 * Umm, how about just "...with norm $$\|y_i\|_V$$ for $$y_i \in V$$ is said..."? Unless a generic placeholder is needed in multiple references, it's really unnecessary here. It's kinda like (perhaps completely analogous to) casting a function argument as  in C.  SamuelRiv (talk) 02:08, 13 December 2011 (UTC)
 * This is actually a very common use of the dot, when one wants to refer to "the norm" (rather than the norm of any particular element). An equivalent notation is $$x\mapsto \|x\|$$, but this is rather clumsy in written prose.   Sławomir Biały  (talk) 14:25, 13 December 2011 (UTC)

Using comments here, the above is the 1st iteration of a possible addition to the list, along with two entries for dot already in the list. Any suggestions for improvement are welcome, as is criticism. : ) --Bob K31416 (talk) 19:00, 14 December 2011 (UTC)


 * This looks ok.  Sławomir Biały  (talk) 19:55, 15 December 2011 (UTC)
 * Thanks. Regards, --Bob K31416 (talk) 01:54, 16 December 2011 (UTC)

Just to reiterate&hellip; the place for this discussion is Wikipedia_talk:WikiProject_Mathematics. That's the best place for content discussions. — Fly by Night  ( talk )  19:30, 15 December 2011 (UTC)
 * Fly by Night, Thanks for the suggestion. I continued here because there was interest and we were making progress. Anyhow, I appreciate your participation and I think the discussion here may be over. Regards, --Bob K31416 (talk) 01:54, 16 December 2011 (UTC)

I added it to the List of mathematical symbols, with a couple of changes: 1) changed readas from "dot" to "(silent)" 2) trimmed explanation. --Bob K31416 (talk) 15:07, 16 December 2011 (UTC)