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I have a function $$\delta:\mathbb{R}\rightarrow [0,1].$$

We obtain this funtion pointwise as follows: For each point $y\in\mathbb{R}$, $\delta(y)$ is a real number in $[0,1]$. More explicitely, $\delta(y)$ can be any number in $[0,1]$ given a specific $y$.

Once $\delta$ is specified for all $y\in\mathbb{R}$, then $\delta$ will have as many elements as ${\mathbb{R}}$. As we are free to chose any $\delta(y)\in[0,1]$ given a specific $y$, $\delta$ can be obtained in uncountably many ways.

The set $\Delta$ is composed of any possible construction of $\delta$.


Below is a discrete example when $y$ and $\delta$ are defined as sets. In the original question $y\in\mathbb{R}$ and $\delta$ was a continuos function on real numbers.


Discrete example: Let $y:=\{0,1,2\}$ and $\delta(y)\in\{0.1,0.5\}$, Then

$\Delta_1=\{0.1,0.1,0.1\}$

$\Delta_2=\{0.1,0.1,0.5\}$

$\Delta_3=\{0.1,0.5,0.1\}$

        ..         .. 

$\Delta_8=\{0.5,0.5,0.5\}$

and we have $\Delta=\{\Delta_1,\Delta_2...,\Delta_8\}$

I must prove that the set $\Delta$ of all possible $\delta(y)\in \Delta$ is convex and compact.

Convexity: For any given $\alpha\in[0,1]$

$$\delta^{'}(y)=\alpha\delta(y)+(1-\alpha)\delta(y)$$ is also a valid function in $\Delta$, therefore $\Delta$ is a convex set.

Compactness: If I can show that for each open cover of $\Delta$ there exists a subcover, then I am done but I dont know how to show it or if there is something simpler.

I will be very grateful for any help,

Thanks in advance.

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    Say again, what is $\Delta$?2012-11-26
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    $\Delta$ is the set of all possible functions $\delta(y)$ due to any arbitrary definition/choice of $u(y)$. For example assume, we are dealing with the discrete case, and we have $u(1)$, $u(2)$ and $u(3)$, $y\in\{1,3\}$. Then I can have $\delta(1)=0.4$, $\delta(2)=0.8$, $\delta(3)=0.3$ and have $\Delta_1=\{0.4,0.8,0.3\}$ or I could have $\delta(2)=0.1$, $\delta(2)=0.1$, $\delta(3)=0.2$ or etc.. and $\Delta:=\{\Delta_1,\Delta_2...\}$2012-11-26
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    Still lost... Is $\Delta$ equal to $[0,1]^\mathbb R$?2012-11-26
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    Each element of $\Delta$ is $[0,1]^{\mathbb{R}}$, because $\delta(y)\in[0,1]^{\mathbb{R}}$. However one can define $u(y)$ as he/she wishes, and therefore the set $\Delta$ contains uncountably many members that are $\in[0,1]^{\mathbb{R}}$2012-11-26
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    You seem to be saying that $\Delta=\{\delta(y)\mid y\in Y\}\subseteq[0,1]^\mathbb R$ for some unspecified set $Y$, and that each $\delta(y)$ is more or less any element of $[0,1]^\mathbb R$ one wants. Then, why should one expect $\Delta$ to be convex (or measurable)?2012-11-26
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    $y\in\mathbb{R}$. The random varibles responsible for creating $\delta(y)$ are also known. They are all possible $u(y)$ creating $\delta(y)$ pointwise and for any specific $y$, $\delta(y)\in[0,1]$. It turns out that $\Delta=\{\delta(y)|y\in[0,1]\}\in[0,1]^{\mathbb{R}}$. It is convex because if I calculate $\alpha\delta+(1-\alpha)(1-\delta)\in\Delta$. It should be also compact. I have this from a paper. It was only mentioning that $\Delta$ was compact with respect to infinity norm. I also didnt understand it.2012-11-26
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    Still lost... :-( but I must leave now, sorry about that.2012-11-26
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    Ok I am sorry. I was not clear enough. I will write it clearer in the evening. Good day to you.2012-11-26
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    I think it would help if you precisely define what you mean by the sentence "arises from the success probability of pointwise Bernoilli distributed random variables u(y) on reals. " I don't know what it means for random variables to be "pointwise Bernoulli distributed".2012-11-26
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    @NateEldredge I edited the question. Do you think it is okay now?2012-11-26
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    No, in fact it seems to be getting worse. First you write $y\in\mathbb R$, then later $y$ is a set. Then you write sets that contain the same element more than once, and it looks like you're trying to make order matter in a set. And it's still not clear (to me) what sort of object $\Delta$ is. In your "discrete example" it appears to be a set of sets of numbers, but further above it sounds like it's supposed to be a set of functions. And what does it mean to say it has $[0,1]^{\mathbb R}$ elements? That's not a cardinal -- do you mean that it has as many elements as $[0,1]^{\mathbb R}$?2012-11-26
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    Once you've clarified what $\Delta$ is, you'll also have to specify a topology on whatever set it's a subset of before you can talk about its compactness.2012-11-26
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    @joriki yes as many elements as $[0,1]^{\mathbb{R}}$. No $y\in\mathbb{R}$. The discrete example is not one to one. It is just for some illustrative purposes. In the question, $y\in\mathbb{R}$ and $\Delta$ is a set of functions, namely all possible, $\delta(y)$. Is $\Delta$ clear now? what topology should I define? I dont know that.2012-11-26
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    No, it's not at all clear. $\delta(y)$ is a function from $\mathbb R$ to $[0,1]$, whereas $[0,1]^{\mathbb R}$ is the set of *all* such functions -- what do you mean when you say that one such function, $\delta (y)$, has as many elements as the set of functions? As far as it makes any sense to consider $\delta$ as a set, it's the set of all pairs $(y,\delta(y))$, and this set has as many elements as $\mathbb R$, not as many as $[0,1]^{\mathbb R}$. Regarding the topology: I have no idea; *you* have to say which topology you're interested in; else compactness has no meaning.2012-11-26
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    $\delta(y)$ is any possible bounded (in $[0,1]$) function on reals $y$ and it has as many as $\mathbb{R}$ number of elements. It is a mistake. I think then the whole set $\Delta$ has as many elements as $[0,1]^{\mathbb{R}}$. This question is coming from a minimax problem. According to the existance of a saddle point for the minimax problem, two sets, one for the minimization and the other for the maximization should be convex and compact. I have two sets and for one I have no problems. The second set is $\Delta$ as I described above. The minimax paper says let $X$ $Y$ be two topological spaces2012-11-26
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    @joriki it doesnt give details. I have the set and I wonder if it is compact such that I will be able to claim that there exits a saddle point.2012-11-26
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    What's the "it" that doesn't give details?2012-11-26
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    @joriki did you receive what I have written to you in chat?2012-11-26

1 Answers 1

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In the end, it seems that $\Delta$ is simply $\Delta=[0,1]^\mathbb R$ the set of all functions defined on $\mathbb R$ with values in $[0,1]$. Then the convexity of $\Delta$ is trivial since any product of convex sets is convex and $\Delta$ is a product of the convex set $[0,1]$. Is this set $\Delta$ compact? For the product topology, this is Tychonoff theorem.

But now I note that later on in the question, you declare that every $\delta$ in $\Delta$ is a continuous function... Thus, $\Delta$ would be $\Delta=C^0(\mathbb R,[0,1])$. Then the convexity of $\Delta$ is trivial. Is this set $\Delta$ compact? This could depend on the topology you put on $\Delta$ (pointwise convergence? uniform convergence?) but in general the answer shall be no.

Note Two confusions seem to plague your understanding:

  • First, functions are not numbers. You cannot at the same time pretend that For each point $y\in\mathbb{R}$, $\delta(y)$ is a real number in $[0,1]$ and that $\delta(y)$ is a function defined on $\mathbb R$. If the first assertion holds, then $\delta$ is a function and $\delta(y)$ is a number. Please watch out for this confusion when you write down your definitions.

  • Second, one does not check convexity the way you check it: if $\delta$ is in a set $\Delta$, then $\delta'=\alpha\delta+(1-\alpha)\delta$ is also in $\Delta$, always! Simply because $\delta'=\delta$. Convexity asks something else, namely that $\alpha\delta_1+(1-\alpha)\delta_2$ is in $\Delta$, for every $\alpha$ in $[0,1]$ and every $\delta_1$ and $\delta_2$ in $\Delta$.

Finally, note that when I declare that the convexity of any product of convex sets is trivial, I mean it. Please write the definitions down and you should see that this is obvious and that the only hypothesis you need is that $[0,1]$ (the target set) is convex. Ditto for the convexity of $C^0(\mathbb R,[0,1])$.

Edit It appears now that $\Delta$ would be the set of continuous nondecreasing functions $\delta$ defined on $\mathbb R$ such that $0\leqslant\delta\leqslant1$, $\delta(y)=0$ for every $y$ small enough $y$ and $\delta(y)=1$ for every $y$ large enough. Then, convexity is still trivial and compactness still dubious (consider the functions $(\delta_t)_{t\in\mathbb R}$ in $\Delta$ defined by $\delta_0:y\mapsto\max\{0,\min\{y,1\}\}$ and, for every $t$, $\delta_t:y\mapsto\delta_0(y-t)$).

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    In the paper he was mentioning that $\Delta$ was compact with respect to the infinity norm. Does it say something to you?2012-11-27
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    Sure, this is the topology of the uniform convergence--then we are probably in the second case of my answer, namely $\Delta=C^0(\mathbb R,[0,1])$. But why should this space be compact when the source set $\mathbb R$ is not, eludes me. (What is the source?)2012-11-27
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    I am sorry, I know one source it was in the matrix movie)) here it is http://arxiv.org/pdf/0707.2926.pdf on page 3, after equation 2.6.2012-11-27
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    The assertion just after equation 2.6 that $\|\delta\|_\infty=\max\limits_{y\in\mathbb R}\delta(y)$ needs some serious justification. Without further hypothesis, this is wrong, consider $\delta(y)=y^2/(y^2+1)$.2012-11-27
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    The highest restriction on $\delta$ is that it is $0$ from $-\infty$ to some real number $y_l$, then increasing from $0$ to $1$ from $y_l$ to $y_u$ and for $y>y_u$, it is one. Is this restriction useful? I want to consider your example but I dont know how)2012-11-27
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    See Edit. $ $ $ $2012-11-28
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    Thank you very much for elaborating in this matter2012-11-28
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    Excuse me. About your previous comment for the infinity norm $||\lambda||_\infty=max_{y\in\mathbb{R}}\delta(y)$, you suggested $\delta(y)=y^2/(y^2+1)$ as a counterexample I guess. Doesn't $\delta(y)=y^2/(y^2+1)$ satisfy $||\lambda||_\infty=max_{y\in\mathbb{R}}\delta(y)$? As long as I know when $y\rightarrow\infty$, both r.h.s. and l.h.s. go to $1$?. I also had a look at your edit. So there is no subsequence which converges for the example you gave. I need to have a compact set to apply the minimax theorem, although i dont know why I need compactness.2013-01-10
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    I will be happy if you could comment on this, since I can not ask it to anybody else. Thanks.2013-01-16
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    Wher $\delta(y)=y^2/(y^2+1)$, the function $\delta$ has a minimum (namely, $0$ reached at $y=0$) and no maximum (since the supremum $1$ is never reached). Let me suggest you check the terms *maximum*, *minimum*, *supremum*, *infimum*.2013-01-16
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    Clear. So the left hand side is $1$ but the right hand side is not because of the term $\max$. There is no maximum element in that set. I guess that $\sup$ would solve the problem but then, it doesnt imply compactness. But how can this be advertised then as a compact set in the paper I showed you? It is strange.2013-01-17