Money is tight

February 2009

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Response to comment on ZFC

Here is a belated response to your question on Talk:Zermelo–Fraenkel set theory.

Regular mathematicians, that is, not logicians, do not think in terms of a metatheory at all. They work "inside" ZFC, which is fine because they are not studying ZFC itself.

In logic, there are two general approaches:

1. Use a strong metatheory and work semantically. For example, we could use ZFC itself as a metatheory for studying ZFC. Then we can do model-theoretic constructions, etc.
2. Use a weak metatheory and work syntactically. We can use Peano arithmetic or some weaker theory of arithmetic as the metatheory. Very often people use PRA because it is a finitistic theory. Now we just study the formal sentences in ZFC, and their relation to each other in terms of provability over the ZFC axioms. There are no models at all.

Usually method 1 is easier because it allows us to work like regular mathematicians and use our intuition, while method 2 has some epistemological benefits because it doesn't presuppose the consistency of strong theories. Actually writing out proofs in in PRA would be a painful exercise, though.

The two approaches complement each other. In many cases, there are general principles that certain model-theoretic techniques can be replaced by syntactic techniques. For example, people usually think of forcing in semantic terms, but there is a general principle that any proof via forcing can be recast into syntactic terms. Similarly, the model-theoretic proof that L is an inner model of ZFC can be recast as a syntactic proof that Con(ZFC) implies Con(ZFC + V=L).— Carl (CBM · talk) 14:06, 5 March 2010 (UTC)

Thank you for your response

Given that f is a correspondence (on a given universal domain of discourse), the idea of "f has a bijective restriction from A to B" - has a very strict unambiguous meaning, which is equivalent to the following definition:

1. A,B are sub-sets of F's universal domain of discourse.

2. For every s,t in B: if a given element in A is corresponding (by f) to s and to t, then s=t.

3. For every s,t in A: if s,t are corresponding (by f) to a given element in B, then s=t.

4. Every element, to which an element in A is corresponding (by f), is in B.

What happens if - due to considerations of symmetry (with the fourth condition) - I add the following condition?

5. Every element, which is corresponding (by f) to an element in B, is in A.

Then I get another definition, very similar to the first one. Yet, the two alternative definitions are not equivalent to each other: For example, the correspondence (on the real numbers): "be the opposite number of" has a bijective restriction from the set of negative numbers to the set of positive numbers - according to both definitions, while the correspondence (on the real numbers) "be the square of" has a bijective restriction from the set of negative numbers to the set of positive numbers - according to the first definition only, yet not according to the second one - which doesn't enable the bijective restriction.

Let's call the first definition (described above): "the definition of classical bijective restriction", and let's call the second definition (received by adding the fifth condition): "the definition of strong bijective restriction". My question is about whether there is a simple brief expression for what I call "strong bijective restriction", or I have to explicitly display the second definition whenever I have to use "strong bijective restrictions".

HOOTmag (talk) 10:58, 15 March 2010 (UTC)

I had a look at the archive 2010_February_23 again. When we say ln(x^2) or any formula, we are being very informal. We do not associate any domain or range to this "function", we just visualize ln(x^2) as an algorithm that spits a real number if we input a real number. Now to talk about a bijection function, we need to specify the function's domain and range. So in the case of ln(x^2), to define a "strong bijective restriction", we would introduce a new function g: Let A be the set of all positive real numbers. Define, for each x in A, g(x)=ln(x^2). Then we show that the function g is bijective.

This works for any function f:X->Y; suppose A is a subset of X, the the restriction of f to A is the function g defined by, for every x in A, g(x)=f(x). Hope that helps. Money is tight (talk) 02:28, 16 March 2010 (UTC)

If f is the function f(x)=x² (on the real numbers), then how can your suggestion help me to declare briefly that f has no strong bijective restriction g from the set A of negative numbers to the set B of positive numbers, when I wouldn't like to use the very term "strong bijective restriction"?

Generally, if f is a given correspondence (which isn't necessarily a function), then how can I state briefly that f has (or doesn't have) strong bijective restrictions from the set A to the set B?

HOOTmag (talk) 08:32, 16 March 2010 (UTC)

I can't think of any brief way to it. Typically in mathematics a long complicated definition is made (for example the free group on a set S) and then we give a name to it (simply calling it the free group on S, when we know the complicated details of the construction in our head). In your case we would define what is meant by a strong bijective correspondence, and keep in mind what the explicit conditions. Money is tight (talk) 03:10, 18 March 2010 (UTC)

Let's put it another way. If $\varphi$ is a formula defined by: $\varphi (x,y)$ ::= x=y², then $\varphi$ induces a function from the set X of positive numbers x to the set Y of negative numbers y. However, $\varphi$ does not strongly induce a function from X to Y, i.e. I can only make sure that every positive x has a negative y such that $\varphi (x,y)$, and that there isn't any other negative y satisfying $\varphi (x,y)$, but I can't make sure that there isn't any other arbitrary y satisfying $\varphi (x,y)$. Here is a counter example - for strongly-inducing a function: The formula $\varphi$, defined by $\varphi (x,y)$ ::= y=x², strongly induces a function from the set X of negative numbers x to the set Y of positive numbers y, namely: not only can I make sure that every negative x has a positive y such that $\varphi (x,y)$ and that there isn't any other positive y satisfying $\varphi (x,y)$, but I can also make sure that there isn't any other arbitrary y satisfying $\varphi (x,y)$.

The notion of a "formula which strongly induces a function" sounds very intuitive to me. Do you still think that there isn't any familiar brief expression for indicating this intuitive notion?

Here's the thing: the formula x=y² does not determine a function. When we have a formula $\varphi (x,y)$ with exactly two free variable, we semantically think of all ordered pairs (x,y) in the domain of discourse in consideration (all real numbers here) that satisfy $\varphi (x,y)$. Now we need to know that no two different ordered pairs can share the same first coordinate for the formula to determine function; otherwise it's just a new relation. So we need the fact that for any x,u,v, if $\varphi (x,u)$ and $\varphi (x,v)$ then u=v. Your second example y=x² does indeed define a function according to the above, whereas your first x=y² doesn't. So "formula which strongly induces a function" should be expressed by the (may not so brief) expression "For any x,u,v, $\varphi (x,u)$ and $\varphi (x,v)$ implies u=v". Money is tight (talk) 02:58, 19 March 2010 (UTC)

Of course, the formula $\varphi$, defined by $\varphi (x,y)$ ::= x=y², does not determine a function in the domain of discourse. However, it does induce (not "strongly" but "regularly") a function from the set X of positive numbers x to the set Y of negative numbers y, because every positive x has a negative y such that $\varphi (x,y)$, and there isn't any other negative y satisfying $\varphi (x,y)$. Agree?

If you do, then how about a "function which strongly induces a bijection"? for example, If $\varphi$ is a formula defined by: $\varphi (x,y)$ ::= y= 2|x|, then $\varphi$ induces a bijection from the set X of negative numbers x to the set Y of positive numbers y. However, $\varphi$ does not strongly induce a bijection from X to Y, i.e. I can only make sure that every negative x has a positive y such that $\varphi (x,y)$, and that there isn't any other (positive) y satisfying $\varphi (x,y)$, and that every positive y has a negative x such that $\varphi (x,y)$, and that there isn't any other negative x satisfying $\varphi (x,y)$, but I can't make sure that there isn't any other arbitrary x satisfying $\varphi (x,y)$. Here is a counter example - for strongly-inducing a bijection: The formula $\varphi$, defined by $\varphi (x,y)$ ::= y=|x|-x, strongly induces a bijection from the set X of negative numbers x to the set Y of positive numbers y, namely: not only can I make sure that every positive y has a negative x such that $\varphi (x,y)$, and that there isn't any other negative x satisfying $\varphi (x,y)$, but I can also make sure that there isn't any other arbitrary x satisfying $\varphi (x,y)$.

The notion of a "formula which strongly induces a bijecion" sounds very intuitive to me. Do you think that there is any familiar brief expression for indicating this intuitive notion?

Weather

Not cold or hot right now? I'd rather be wherever you are. It's abnormally hot down here in Gippsland. Uncomfortably hot. This is TMI, but right now, I'm sitting naked at my computer, with a strong fan at my back cooling me down. -- Jack of Oz ... speak! ... 07:26, 18 March 2010 (UTC)

Is it also humid there? HOOTmag (talk) 08:24, 18 March 2010 (UTC)

Yes. I've known it to be worse, but it's sticky enough. -- Jack of Oz ... speak! ... 07:37, 19 March 2010 (UTC)

Where in Gippsland? Thorpdale? Stratford? Port Whelshpool? HOOTmag (talk) 11:00, 19 March 2010 (UTC)

Wikipedia:Reference desk/Mathematics

I responded to your comment/question at the mathematics reference desk regarding "life after undergraduate education". Cheers, PST 11:44, 24 August 2010 (UTC)

tb

Hello, Money is tight. You have new messages at Toby Bartels's talk page.
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Advice re multiple accounts

Hello Money is tight, a checkuser has determined that you are operating two user accounts, this one and User:Gud music only. Please carefully read our policy on sock puppetry and ensure that you comply with it fully. In particular note that you should show on both user pages that you are the same editor to avoid confusing others, and you must avoid any appearance that the two accounts support each other in discussions and edits. If you have further questions you can ask them here and I will try to answer. Franamax (talk) 01:35, 22 January 2011 (UTC)

Hormone question

Here's my amateur medical advice: regular, vigorous exercise can help take the edge off the libido. Occasional masturbation helps too! thx1138 (talk) 23:01, 15 March 2011 (UTC)

Non commercial images

Hi, Although it may be legal to upload a non-commercial personal use image to Wikipedia, it is not within the policy for Wikipedia. The images here should be free for all kinds of uses, including commercial and derivatives. Rather than copying if the diagram was drawn again by someone else and uploaded with a free license it would be acceptable. It could use different line thickness, length, arrows, letter placement or even different symbols. I thought it may be too simple for copyright, but it is probably over the threshold of originality. Graeme Bartlett (talk) 20:34, 17 May 2011 (UTC)

References for uniqueness of injective resolution

Hi, thanks for your response at the helpdesk. Do you happen to know a text I could look at to cite for the uniqueness of injective resolutions? Please let me know on my talkpage. Thanks! Rschwieb (talk) 14:05, 28 February 2012 (UTC)

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Reference desk question removed

Hi there. Your question on the Science Reference Desk about drugs to inhibit anxiety has been removed, as the reference desk guidelines at WP:RD/G/M do not permit us to give any kind of medical diagnosis, prognosis, or treatment recommendation. If you would like to discuss the removal of this content, please comment at the Reference Desk talk page. Thanks. Tevildo (talk) 22:01, 14 April 2016 (UTC)

Private message

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January 2019

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Thank you. Robert McClenon (talk) 20:47, 4 January 2019 (UTC)

Responded to your question

I Responded to your question at the science ref desk. It took a little time to dig up the references, so I had to bury it invisibly into the archives at Wikipedia:Reference desk/Archives/Science/2019 January 4. This is just a note in case you are still interested. It follows my general view of everything that everybody usually gets everything backwards. ;-)John Z (talk) 02:48, 16 January 2019 (UTC)