# Tag Info

34

I sent Professor Purdy an email. I asked him what he recalled about the incident. With his permission I've copied his correspondence below. Dear Jacob, Yes, I was there, and I'm the one who told the story to Paul Hoffman, who then included it in his book "The man who loved only numbers." The 30 page proof was written by Jack Bryan just ...

30

Let me ask you a dual question: I am a mathematics student in set theory, why don't category theory students do set theory? I find it strange that most books on category theory have only a naive handling of set theory. Now let me answer your question. Category theory is an impressive tool for abstraction, but analysis is not always in need for abstraction - ...

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Yes, you are right. It was Jack Bryant, not Jack Brian. He might have retired by now. By the way, even before Erdos came to town, it was generally agreed that there must be a proof that was shorter than 30 pages, but not a two liner! Professor Don Allen talked to Erdos the next day to see if he could help with the research problem that had generated this ...

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Section 1 Let us begin with the following theorem. Theorem 1 Let $G$ be a topological group. If $G$ admits a Lie group structure, then this structure is unique up to diffeomorphism. Proof: Suppose that $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ are smooth structures (maximal smooth atlases) on $G$ that make it a Lie group. Observe that the ...

15

There are two questions here, in reality, I think. First, in brief, I am told by many people that I "do functional analysis in the theory of automorphic forms", and I certainly do find a categorical viewpoint very useful. Second, in brief, it is my impression that the personality-types of many people who'd style themselves "(functional) analysts" might be ...

15

The first result that you stated is commonly known as the Gelfand-Naimark-Segal Theorem. It is true for arbitrary C*-algebras, and its proof employs a technique known as the GNS-construction. This technique basically allows one to construct a Hilbert space $\mathcal{H}$ from a given C*-algebra $\mathcal{A}$ such that $\mathcal{A}$ can be isometrically ...

15

I believe that one of the standard proofs works. Let $F(x) := \intop_a^x f^\prime (t) dt$. Then $F$ is differentiable and its derivative is $f^\prime$ due to a standard estimate that has nothing to do with AC. $(F-f)^\prime = 0$, hence it is constant. This boils down to the one-dimensional case: just consider $g := \Vert F-f-F(a)+f(a) \Vert$. It is a ...

14

The existence of a Hamel basis for $\ell^p$ cannot be proved without some of the axiom of choice, which in modern terms usually means that we cannot write it explicitly. It is consistent with ZF+DC (a weak form of the axiom of choice which is sufficient to do a lot of the usual mathematics) that all sets of real numbers have Baire property, and in such ...

14

The set $C$ is the image of the compact set $X = [-1,1]^{\mathbb N}$ (with the product topology) under the map $F: X \to \ell^2$, where $F(x)_j = x_j a_j$. Since the image of a compact Hausdorff space under a continuous map is compact, we just need to show that $F$ is continuous. Note that $\Vert F(x) - F(y)\Vert^2 = \sum_{j=1}^\infty (x_j - y_j)^2 a_j^2$. ...

13

Let $V$ be a vector space over the field $\mathbb{F}$. A norm $$\| \cdot \|: V \longrightarrow \mathbb{F}$$ on $V$ satisfies the homogeneity condition $$\|ax\| = |a| \cdot \|x\|$$ for all $a \in \mathbb{F}$ and $x \in V$. So the metric $$d: V \times V \longrightarrow \mathbb{F},$$ $$d(x,y) = \|x - y\|$$ defined by the norm is such that $$d(ax,ay) = \|ax - ... 12 The answer is yes. In Chapter VI § 2 of Selected Topics in Infinite Dimensional Topology (Bessaga & Pełczyński) is stated the surprising result that \mathbb{R}^\infty, the Hilbert space \ell_2 and the unit sphere S \subset\ell_2 are homeomorphic spaces ({}^*\!). The background required (developed in the previous chapters) seems to be rather ... 11 Without assuming additional properties on S and T, such as commutation or self-adjointness (on a Hilbert space) — see points 2. and 3. at the end of the answer — continuity of the spectrum as a map from \mathcal{L}(X) to the compact subsets of \mathbb{C} with the Hausdorff distance d_H doesn't hold, because the spectrum can ... 11 There are several ways of doing this, but I'll go with the most "elementary". Let \varphi be a nonzero multiplicative functional on \ell^\infty(\mathbb{N}). Since \varphi(1)=\varphi(1^2)=\varphi(1)^2, we get that \varphi(1)=1 (it cannot be zero, because then \varphi=0). Now let a\in\ell^\infty(\mathbb{N}) such that a(n)\in\{0,1\} for all ... 10 As discussed in the comments, the actual question is \sigma-additivity of the limit of a Cauchy sequence of complex measures. If you're only interested in this part you can jump to the claim towards the end of the answer, but for the sake of completeness I'll give the definitions and the entire argument that the space of complex measures of bounded ... 10 Originally mathematics was intended to describe the real world. We then continued to develop it using the intuition of how the real world behaves in order to describe how mathematical objects would behave. In the 19th and 20th century mathematics had several foundational crises. It turned out that intuition is not a good enough foundation for mathematics. ... 10 To answer your 4 questions briefly: Your \alpha is an operator from a space of real valued functions to itself. You probably require the functions to be bijective in some sense (in order for f^{-1} to even make sense), perhaps with both domain and codomain a subset of the reals. By the recursive relation, each f_n(x) is also bijective in the same ... 10 The chain rule for functional differentiation is just the continuum generalisation of the usual chain rule for differentiation of a function of many variables f(y_1,y_2,\ldots,y_N) = f(\mathbf{y}), which reads$$ \frac{\partial f(\mathbf{y})}{\partial x_i(\mathbf{y})} = \sum\limits_{j=1}^N\frac{\partial y_j}{\partial x_i}\frac{\partial f}{\partial y_j}.  ...

9

No, we cannot conclude that the operator is trace class. For example, let a Hilbert space have orthonormal basis $e_1,f_2,e_2,f_2,e_3,f_3,\ldots$, and $T$ interchanges $e_i,f_i$, while multiplying both by a positive real $\lambda_i$. That is, in these coordinates, the matrix of $T$ is a list of diagonal blocks, with the $i$-th diagonal block being ...

9

People aren't going to switch to a new formalism unless they have a compelling reason to. Everyone already knows measure theory (including the professors teaching analysis courses). It is very powerful and suffices for many applications, so until someone convinces the mathematical community that an alternate theory of integration would fix their troubles, ...

9

Just a partial answer... As Qiaochu remarks, the "standard viewpoint" will not change without considerable impetus. The fact that some variant of "Lebesgue" integration is not "perfect" doesn't matter: it is "good enough". Further, I would claim that, in fact, "the integral" people mostly use is not so much formally defined by any particular set-up, but is ...

9

The following excerpt is from Measure Theory Vol 2 by Vladimir Bogachev: In the middle of the 20th century there was a very widespread point of view in favor of presentation of the theory of integration following Daniell’s approach, and some authors even declared the traditional presentation to be “obsolete”. Apart the above-mentioned conveniences ...

9

In what follows, $\mathbb{F}$ shall denote either $\mathbb{R}$ or $\mathbb{C}$. The standard Euclidean topology on $\mathbb{F}$ shall be denoted by $\tau_{\mathbb{F}}$. It is important to know that the weak topology on a vector space $V$ over $\mathbb{F}$ can only be defined after we have already endowed $V$ with a linear topology, i.e., a ...

9

This is an admittedly confusing abuse of terminology which your book appears to make even more confusing by using somewhat nonstandard terminology. The following is more standard but also confusing in its own way. For what follows let $X$ and $Y$ be Banach spaces. A bounded linear operator from $X$ to $Y$ is a function $T: X \to Y$ which is linear and ...

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You are quite right to ask for the context of these definitions. One place to look in this case is http://www-history.mcs.st-and.ac.uk/HistTopics/Topology_in_mathematics.html There are three reasons for abstraction: To cover many known examples. To simplify proofs by giving the key reasons why something is true. To be available for new examples. ...

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What Lax shows is that there is a finitely additive, rotationally invariant set function $m \colon P(S^1) \to [0,1]$ on the circle such that $m(S^1) = 1$: This implies in particular that it is impossible to decompose $S^1$ paradoxically into a disjoint union of finitely many pieces $A_1,\dots,A_n$ in such a way that $S^1$ can be written as disjoint union ...

9

James's theorem asserts the following: if every continuous linear functional on $E$ attains its norm then $E$ is reflexive. This means that on every non-reflexive space we can find a codimension 1 subspace for which the quotient norm needs the infimum in place of a minimum. In the comments there were already a few links to threads discussing explicit ...

9

Hints: $\mathcal O(n)$ is the continuous inverse image of a closed set… There exists a rather obvious bound for every element in $\mathcal O(n)$ and thus this set is bounded… Now Heine-Borel and we're done. Added: Thanks to the comments by Berci I think some confusion may happen here, since the orthogonal group is not the inverse image of $\,\{1,-1\}\,$ ...

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