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November 4

Complete vector spaces under the uniform norm

Hi all:

I'm wondering whether the following spaces are complete under the uniform norm $\|f(x)\|_{\infty }=\sup _{x}(|f(x)|)$ , and I'm a little concerned my arguments are too rough and I'm neglecting things which could lead me to be wrong because i haven't thought it through well enough, I'd appreciate it if someone could point out areas where I'm going wrong, which I have no doubt there will be!

i) The space $C_{b}(\mathbb {R} )$ of bounded continuous functions $f:\,\mathbb {R} \to \mathbb {R}$ .

ii) The space $C_{0}(\mathbb {R} )$ of continuous functions $f:\,\mathbb {R} \to \mathbb {R}$ with $f(x)\to 0$ as $|x|\to \infty$

iii) The space $C_{c}(\mathbb {R} )$ of continuous functions $f:\,\mathbb {R} \to \mathbb {R}$ with $f(x)=0$ for $|x|$ sufficiently large.

My thoughts:

i) I know that on the bounded interval ( $f:\,[0,1]\to \mathbb {R} )$ , the space is complete, and so presumably the same is true of any interval [a,b] by stretching and translating. The functions f_n (Cauchy sequence) converge to their pointwise limit on these intervals, and since the interval is arbitrary, the functions have a limit f which is continuous on any interval [a,b] and thus bounded on it too. Thus, the only manner in which f can be unbounded is as x tends to (± infinity): however, if this is the case (WLOG take + infinity), then since f is the pointwise limit of the f_n, and the sequence is Cauchy which implies that, if as n tends to infinity the f_n have arbitrarily increasing upper bounds (pointwise convergence to an unbounded function) and we can make the f_n arbitrarily close for large enough n > some N by the Cauchy property, the f_n all tend to infinity as x tends to infinity, since they must remain as close to each other as we require by choice of N. Thus the space is complete.

ii) The functions are continuous and by their definition must be bounded, so by (i) they converge to a bounded continuous function in the limit, and because this function is in fact their pointwise limit, if it does not tend to 0 as |x| tends to 0, then the f_n do not either, and so by a contradiction argument, f too must tend to 0, and so the space is complete again.

iii) I believe I've constructed a series of functions which are 'truncations' of f(x)=1 if |x|<1, 1/x if |x|>1 and which in the limit tend to this f(x) which is not equal to 0 for sufficiently large |x|, but the sequence is Cauchy because as m,n tend to infinity, f_m and f_n agree with each other precisely, for larger and larger values of |x|, and thus $\|\cdot \|_{\infty }$ tends to 0, so the seq. is Cauchy and hence this is not a complete space.

Sorry if any of that doesn't make sense, please try to muddle through :) I do intend to refine this argument more when I write the question up in full, but since it's quite long I'd rather sort things out in rough before I dive in fully rigorously - I have, no doubt, made some stupid mistakes, and your help would be greatly appreciated! (And thanks for reading this far, too.) Typeships17 (talk) 02:50, 4 November 2009 (UTC)[reply]

i) This looks right but I think you can tighten it up a bit. To show the pointwise limit f is bounded, pick any ε > 0. There exists an N such that ||f_N - f_n|| < ε for all n > N. Suppose f_N is bounded by M, then you can show f is bounded by M + ε.

ii) It's not true in general that the pointwise limit of a sequence of functions that all go to 0 as x goes to infinity must go to 0 as x goes to infinity. For example let g_n(x) = 1 for |x| < n and g_n(x) = 0 otherwise. However for any ε > 0 you can find N as before, and then some B such that |f_N(x)| < ε for all |x| > B, and then you should be able to argue that |f(x)| < 2ε for all |x| > B.

iii) This looks good.

Rckrone (talk) 04:09, 4 November 2009 (UTC)[reply]

Wow, I wasn't expecting an answer so quickly, thanks very much! :D I'll get to writing them up properly then! Typeships17 (talk) 04:16, 4 November 2009 (UTC)[reply]

As a further remark, note that the space (iii) is dense in the space (ii), and different from it, so it's not complete. Both (i) and (ii) are closed subspaces of the space B(R) of all bounded functions R→R, so (i) and (ii) are complete since the latter is complete. For analogous cases, you may find it useful the general result (same proof) that for any set S and any Banach E the space B(S,E) with the sup norm is still a Banach space (this way you get at once the completeness of a lot of norms based on the sup norm: dual and operator norms, &c). Although (iii) is not complete with the sup norm, it has another natural structure of TVS as inductive limit (in the category TVS) of the inclusions C(K)→C(R), K compact. This topology is locally convex, non metrizable, sequentially complete. --pma (talk) 08:28, 4 November 2009 (UTC)[reply]

Two parallelogram areas theorem

When two parallelograms lie between two parallels and same base, their areas are equal. I have got problem with the proof. When the figures are like this, it can be done by proving the triangles congruent (Side-angle-angle) but when the figures are like this, how do we do it? Srin ivas 15:32, 4 November 2009 (UTC)[reply]

What about adding congruent triangles to both, so as to obtain congruent trapezoids. --pma (talk) 15:53, 4 November 2009 (UTC)[reply]

(e/c) If all else fails, you can introduce sufficiently many other parallelograms in between so that the first case applies to each pair of neighbours, and use transitivity of equality. — Emil J. 15:57, 4 November 2009 (UTC)[reply]

You can also make the same congruent triangle argument with triangles ACE and BDF. You just need to do some subtraction to get to the areas you want. Rckrone (talk) 18:56, 4 November 2009 (UTC)[reply]

Incompleteness theorem

The incompleteness theorem says that in any mathematical system, there are statements which can be neither proved nor disproved. Is there any way to prove that a statement is unprovable by this theorem? --75.50.49.177 (talk) 23:05, 4 November 2009 (UTC)[reply]

I'm afraid I don't know what you're really asking. You'll probably need to phrase it more precisely before anyone can help. In the meantime you might take a look at our article on Gödel's incompleteness theorems and see if it addresses your question. --Trovatore (talk) 23:12, 4 November 2009 (UTC)[reply]

The incompleteness theorem states that some statements are neither provable nor disprovable. However, is there any way of determining which statements those are? --75.50.49.177 (talk) 23:14, 4 November 2009 (UTC)[reply]

For a given theory T satisfying the hypotheses, the (modified) proof of the first incompleteness theorem gives an effective way of finding a sentence G_T such that, if T is consistent, then G_T is true, yet neither provable nor refutable from T. Sorry I can't make it less complicated than that. All the stipulations are necessary in order to have a correct statement of the situation, so you'll just have to work your way through them. -Trovatore (talk) 23:21, 4 November 2009 (UTC)[reply]

That says how to construct a statement which is unprovable. However, given a statement (e.g. the Riemann Hypothesis), is it possible to determine whether or not that statement is provable? --70.250.215.137 (talk) 02:41, 5 November 2009 (UTC)[reply]

Not in general, no. Well, at least not by any fixed mechanical procedure. In technical terms, the set of theorems of T is computably enumerable but not computable. --Trovatore (talk) 02:44, 5 November 2009 (UTC)[reply]

As Trovatore says, not in general, no. But there are some specific examples where undecidability can be proved: one of the most famous problems of the early 20th century was to prove or disprove the continuum hypothesis. This turns out to be both unproveable and un-disprovable (also known as independence) if you start from the usual axioms of set theory. The proof of independence was done in the 1960's by Paul Cohen and doesn't use the incompleteness theorem directly (it uses a complicated technique called forcing). That proof was considered a very major result. For other examples, see the article about independence that I linked to. 69.228.171.150 (talk) 05:39, 5 November 2009 (UTC)[reply]

There are plenty of mathematical systems in which every statement can be either proved or disproved. Examples include the theory of real closed ordered fields and the theory of dense linear orderings without endpoints. There are several hypotheses in Gödel's incompleteness theorem and they are all important for the result. — Carl (CBM · talk) 03:28, 5 November 2009 (UTC)[reply]

Hmm, it occurs to me that I gave the answer sbout the theorems being c.e. but not computable without a proof immediately in mind, and when I thought of one, it requires T to be Sigma-1 sound. (The proof is, you can reduce the halting problem to the theorems of T -- given a Turing machine that you want to know whether it halts, just decide whether T proves that it halts. The usual hypotheses of the Gödel theorems suffice to show that, if the TM halts, then T proves that it halts, but for the converse you need that T is Sigma-1 sound. Does anyone see how to eliminate this hypothesis? I flat don't believe that the theorems of, say, PA+~Con(PA) form a computable set, even though that theory proves that certain Turing machines halt that actually don't halt.) --Trovatore (talk) 05:24, 5 November 2009 (UTC)[reply]

As you're saying, no consistent completion of PA is computable, whether that completion is sound or not. Syntactically, you get rid of the soundness assumption by using Rosser's trick. But there is also a completely computational proof, related to the concept of a PA degree. We can prove that given any completion T of PA and any nonempty

\Pi _{1}^{0}

class C in Cantor space 2^ω, there is an element of C computable from T. So if you take T to be a

\Pi _{1}^{0}

class with no computable elements, you obtain a corollary that there is no computable completion of PA. — Carl (CBM · talk) 11:25, 5 November 2009 (UTC)[reply]

Well, unless I'm missing something, that's a little different question. I wasn't asking about completions. The question is, given a consistent r.e. theory extending PA and closed under logical consequence, can you show that that theory is not computable? --Trovatore (talk) 11:34, 5 November 2009 (UTC)[reply]

Yes, and asking it to extend Q rather than PA is more than enough. — Emil J. 11:39, 5 November 2009 (UTC)[reply]

Rosser's trick is one way to prove it, and the other argument by Carl works too: you just have to observe that every decidable theory has a decidable completion. The usual way to prove the theorem along the lines of your proof above is to take a recursively inseparable pair of r.e. sets instead of the halting problem. — Emil J. 11:44, 5 November 2009 (UTC)[reply]

Why does every decidable theory have a decidable completion? --Trovatore (talk) 20:02, 5 November 2009 (UTC)[reply]

You can do this explicitly. The language is countable, so fix an enumeration of all sentences. Then go along the list, and for each sentence φ use your decision procedure to determine which of φ and ~φ is provable from your theory so far. If one of them is, add it to the theory, if neither is, add φ. This gives a completion which is decidable by construction (since you can decide a sentence just by repeating the construction) Thus all you have to show is that if T is decidable, then T union {φ} is decidable, and this is obvious (to see if θ follows from T u {φ}, check if φ→θ follows from T). Algebraist 21:39, 5 November 2009 (UTC)[reply]

Nice; thanks. --Trovatore (talk) 22:19, 5 November 2009 (UTC)[reply]

Exactly. I think I did misread the original question. For PA (or any consistent extension of Q) the set A of formulas φ such that

{\mbox{PA}}\vdash \varphi

and the set B of formulas θ such that

{\mbox{PA}}\vdash \lnot \theta

are already recursively inseparable r.e. sets, that is, there is no computable set C including A and disjoint from B. Any consistent decidable extension of PA would give such a set, so there is no such extension. So PA (and Q) is "essentially undecidable". — Carl (CBM · talk) 12:48, 5 November 2009 (UTC)[reply]