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  1. MHB Master
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    #1
    I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

    I am focused on Chapter 1: Continuity ... ...

    I need help with an aspect of the proof of Theorem 1.8.17 ... ...

    Duistermaat and Kolk's Theorem 1.8.17 and its proof (including the preceding relevant definition) read as follows:






    Towards the end of the above proof, D&K write the following:

    " ... ... In order to demonstrate that $ \displaystyle K$ is closed, we prove that $ \displaystyle \mathbb{R}^n \text{\\}K$ is open. Indeed, choose $ \displaystyle y \notin K$ and define $ \displaystyle O_j = \{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \gt \frac{1}{j} \}$ for $ \displaystyle j \in \mathbb{N}$. ... ... "


    D&K go on to describe the union of the $ \displaystyle O_j$ as an open cover, implying, of course, the the sets $ \displaystyle O_j$ are open ... BUT ... why/how are these sets open?


    Hope someone can help ...

    Peter
    Last edited by Peter; February 11th, 2018 at 02:28.

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    #2
    Hi, Peter.

    Quote Originally Posted by Peter View Post
    D&K go on to describe the union of the $ \displaystyle O_j$ as an open cover, implying, of course, the the sets $ \displaystyle O_j$ are open ... BUT ... why/how are these sets open?
    Peter
    There are two nice ways I can think of that are both good techniques/tools to be familiar with.

    1) Recall from your previous post that the norm function is continuous on $\mathbb{R}^{n}$. Note that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. Since the preimage of an open set with respect to a continuous function is open, $O_{j}$ is open. (Note: This is a common trick in analysis for proving a set is open; i.e., show it is the preimage of an open set of a continuous function).

    2) Method 2 is much more geometric in nature (Note: Drawing a picture in 2-dimensions is very useful here). We want to show that every point $p\in O_{j}$ has an open ball about it that does not intersect the set $\{x:\|x-y\|\leq 1/j\}.$ By choosing $\epsilon$ so that $0<\epsilon<\|p-y\|-1/j$ (which we can do because $\|p-y\|>1/j$), the triangle inequality can be used to show that the ball $\{z:\|z-p\|<\epsilon\}$ and the set $\{x:\|x-y\|\leq 1/j\}$ do not intersect (see if you can use the triangle inequality to prove that for any $z$ in the ball we have $\|z-y\|>1/j$ thus proving the two are disjoint).
    Last edited by GJA; February 11th, 2018 at 02:56.

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    #3 Thread Author
    Quote Originally Posted by GJA View Post
    Hi, Peter.



    There are two nice ways I can think of that are both good techniques/tools to be familiar with.

    1) Recall from your previous post that the norm function is continuous on $\mathbb{R}^{n}$. Note that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. Since the preimage of an open set with respect to a continuous function is open, $O_{j}$ is open. (Note: This is a common trick in analysis for proving a set is open; i.e., show it is the preimage of an open set of a continuous function).

    2) Method 2 is much more geometric in nature (Note: Drawing a picture in 2-dimensions is very useful here). We want to show that every point $p\in O_{j}$ has an open ball about it that does not intersect the set $\{x:\|x-y\|\leq 1/j\}.$ By choosing $\epsilon$ so that $0<\epsilon<\|p-y\|-1/j$ (which we can do because $\|p-y\|>1/j$), the triangle inequality can be used to show that the ball $\{z:\|z-p\|<\epsilon\}$ and the set $\{x:\|x-y\|\leq 1/j\}$ do not intersect (see if you can use the triangle inequality to prove that for any $z$ in the ball we have $\|z-y\|>1/j$ thus proving the two are disjoint).

    Thanks for the help, GJA ...

    Just now reflecting on what you have written ...

    Thanks again,

    Peter

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    #4 Thread Author
    Quote Originally Posted by GJA View Post
    Hi, Peter.



    There are two nice ways I can think of that are both good techniques/tools to be familiar with.

    1) Recall from your previous post that the norm function is continuous on $\mathbb{R}^{n}$. Note that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. Since the preimage of an open set with respect to a continuous function is open, $O_{j}$ is open. (Note: This is a common trick in analysis for proving a set is open; i.e., show it is the preimage of an open set of a continuous function).

    2) Method 2 is much more geometric in nature (Note: Drawing a picture in 2-dimensions is very useful here). We want to show that every point $p\in O_{j}$ has an open ball about it that does not intersect the set $\{x:\|x-y\|\leq 1/j\}.$ By choosing $\epsilon$ so that $0<\epsilon<\|p-y\|-1/j$ (which we can do because $\|p-y\|>1/j$), the triangle inequality can be used to show that the ball $\{z:\|z-p\|<\epsilon\}$ and the set $\{x:\|x-y\|\leq 1/j\}$ do not intersect (see if you can use the triangle inequality to prove that for any $z$ in the ball we have $\|z-y\|>1/j$ thus proving the two are disjoint).



    Hi GJA,

    Thanks again for your extremely helpful post ...

    I get the general idea of what you are saying in 1) ... but can you clarify some details ...

    You write:

    " ... ... Note that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. ... ... "


    How did you "see" that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. ... ...?

    Indeed, further, how would you rigorously demonstrate that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. ... ...?



    Hope you can help further ...

    Peter

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    #5 Thread Author
    Quote Originally Posted by GJA View Post
    Hi, Peter.



    There are two nice ways I can think of that are both good techniques/tools to be familiar with.

    1) Recall from your previous post that the norm function is continuous on $\mathbb{R}^{n}$. Note that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. Since the preimage of an open set with respect to a continuous function is open, $O_{j}$ is open. (Note: This is a common trick in analysis for proving a set is open; i.e., show it is the preimage of an open set of a continuous function).

    2) Method 2 is much more geometric in nature (Note: Drawing a picture in 2-dimensions is very useful here). We want to show that every point $p\in O_{j}$ has an open ball about it that does not intersect the set $\{x:\|x-y\|\leq 1/j\}.$ By choosing $\epsilon$ so that $0<\epsilon<\|p-y\|-1/j$ (which we can do because $\|p-y\|>1/j$), the triangle inequality can be used to show that the ball $\{z:\|z-p\|<\epsilon\}$ and the set $\{x:\|x-y\|\leq 1/j\}$ do not intersect (see if you can use the triangle inequality to prove that for any $z$ in the ball we have $\|z-y\|>1/j$ thus proving the two are disjoint).


    Just been reflecting on 2) ... and need your further help ...

    You write:

    " ... ... the triangle inequality can be used to show that the ball $\{z:\|z-p\|<\epsilon\}$ and the set $\{x:\|x-y\|\leq 1/j\}$ do not intersect ... "


    Can you help me to demonstrate the above ... ?


    Would be grateful for some help ... ...

    Peter

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    #6
    Hi, Peter.

    Quote Originally Posted by Peter View Post
    You write:

    " ... ... Note that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. ... ... "

    How did you "see" that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. ... ...?
    I will do my best to outline where this methodology/thought process comes from. I apologize in advance if I seem to ramble a bit in places. The first section of the response outlines the thought process from a general point of view. The second section then applies the ideas to your specific question.

    Section 1:

    • The first and most important point bar none is that someone showed/taught me this method (as I am trying to show you now). It's not nearly as difficult to find a hidden treasure when you (basically) know where it's buried. I cannot stress this point enough. Furthermore, I hope it resonates and reassures you that you are doing the right thing by asking questions. I also hope it helps take some of the pressure off your own studies to know that you're seeing the application of an idea, not the invention of it.
    • Once you know to think along these lines, trying to reverse-engineering the reasoning behind them is a great exercise (and is why your question is a good one). Perhaps it would look something like this:

    Q: What is it we want to show?
    A: We want to show this set is open.
    Q: Is there a theorem or definition we know that tells us when a set is open?
    A: Hmm...Aha! We know that, by definition, a continuous function from one space to another is one with the property that the preimage of any open set in the codomain space is open in the domain space.
    Q: How can we actually apply this idea when the time comes?
    A: We can try to recognize the defining characteristic of the set we are interested in (in this case $O_{j}$) as a restriction on the range of a certain function (which we must identify). If we are able to do this, then the set we are interested in is, by definition, the pre-image of the restricted range of the function. Note: This is in some sense akin to computing definite integrals using the Fundamental Theorem of Calculus. You want to try to view the integrand as the derivative of some function (which you must figure out). IF you can find this (so-called) antiderivative, the theorem tells you how to calculate the original integral.

    This method of seeing a new technique, then trying to reverse-engineer the reasoning behind it is one of the best ways to truly internalize new ideas. Creating a sequence of smaller leading questions to get to a final question whose answer you know (i.e., the new method you've been introduced to) can be quite helpful. If you're familiar with the film, think Tom Cruise in the final scene of "A Few Good Men." He knows the answer his witness wants to give, he just needs to ask the right sequence of smaller leading questions before the witness will say the thing that ties the entire case together.

    Section 2:
    • Being primed to think in terms of pre-images of continuous functions, the primary step to solving the problem regarding the openness of $O_{j}$ is converted into a question of whether we can view the defining characteristic of points in the set $O_{j}$ as some sort of restriction on a function (as mentioned above, this is like you knowing the Fundamental Theorem of Calculus already and you just need to know if you can think of the integrand as the derivative of some function or not). The condition defining $O_{j}$ is the set of all $x\in\mathbb{R}^{n}$, such that

    $\|x-y\|>1/j,$

    where $y$ is some fixed element of $\mathbb{R}^{n}$. In words, the inequality is saying that the real number $\|x-y\|$ must be strictly larger than $1/j.$ This leads us to consider the interval $(1/j,\infty).$ Now, how did we go from thinking of $x$ and $y$ as being elements of $\mathbb{R}^{n}$ to thinking about the real number $\|x-y\|$? We must have applied a function from $\mathbb{R}^{n}$ to $\mathbb{R}$. This is the norm function, which we knew to be continuous from your previous post.

    Quote Originally Posted by Peter View Post
    Indeed, further, how would you rigorously demonstrate that $O_{j}$ is the preimage of the open subset $(1/j,\infty)\subseteq\mathbb{R}$ under the function $f_{y}(x)=\|x-y\|$. ... ...?
    We have $f_{y}^{-1}\Big((1/j,\infty)\Big)=\{x\in\mathbb{R}^{n}:f_{y}(x)>1/j\}=\{x\in\mathbb{R}^{n}:\|x-y\|>1/j\}=O_{j},$ where the first equality is the definition of pre-image, the second is the definition of $f_{y}(x)$, and the third is the definition of $O_{j}$.

    I hope this post was coherent enough to make sense of and that it helps answer your question of how to "see" the answer.

  7. MHB Master
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    #7 Thread Author
    Quote Originally Posted by GJA View Post
    Hi, Peter.



    I will do my best to outline where this methodology/thought process comes from. I apologize in advance if I seem to ramble a bit in places. The first section of the response outlines the thought process from a general point of view. The second section then applies the ideas to your specific question.

    Section 1:

    • The first and most important point bar none is that someone showed/taught me this method (as I am trying to show you now). It's not nearly as difficult to find a hidden treasure when you (basically) know where it's buried. I cannot stress this point enough. Furthermore, I hope it resonates and reassures you that you are doing the right thing by asking questions. I also hope it helps take some of the pressure off your own studies to know that you're seeing the application of an idea, not the invention of it.
    • Once you know to think along these lines, trying to reverse-engineering the reasoning behind them is a great exercise (and is why your question is a good one). Perhaps it would look something like this:

    Q: What is it we want to show?
    A: We want to show this set is open.
    Q: Is there a theorem or definition we know that tells us when a set is open?
    A: Hmm...Aha! We know that, by definition, a continuous function from one space to another is one with the property that the preimage of any open set in the codomain space is open in the domain space.
    Q: How can we actually apply this idea when the time comes?
    A: We can try to recognize the defining characteristic of the set we are interested in (in this case $O_{j}$) as a restriction on the range of a certain function (which we must identify). If we are able to do this, then the set we are interested in is, by definition, the pre-image of the restricted range of the function. Note: This is in some sense akin to computing definite integrals using the Fundamental Theorem of Calculus. You want to try to view the integrand as the derivative of some function (which you must figure out). IF you can find this (so-called) antiderivative, the theorem tells you how to calculate the original integral.

    This method of seeing a new technique, then trying to reverse-engineer the reasoning behind it is one of the best ways to truly internalize new ideas. Creating a sequence of smaller leading questions to get to a final question whose answer you know (i.e., the new method you've been introduced to) can be quite helpful. If you're familiar with the film, think Tom Cruise in the final scene of "A Few Good Men." He knows the answer his witness wants to give, he just needs to ask the right sequence of smaller leading questions before the witness will say the thing that ties the entire case together.

    Section 2:
    • Being primed to think in terms of pre-images of continuous functions, the primary step to solving the problem regarding the openness of $O_{j}$ is converted into a question of whether we can view the defining characteristic of points in the set $O_{j}$ as some sort of restriction on a function (as mentioned above, this is like you knowing the Fundamental Theorem of Calculus already and you just need to know if you can think of the integrand as the derivative of some function or not). The condition defining $O_{j}$ is the set of all $x\in\mathbb{R}^{n}$, such that

    $\|x-y\|>1/j,$

    where $y$ is some fixed element of $\mathbb{R}^{n}$. In words, the inequality is saying that the real number $\|x-y\|$ must be strictly larger than $1/j.$ This leads us to consider the interval $(1/j,\infty).$ Now, how did we go from thinking of $x$ and $y$ as being elements of $\mathbb{R}^{n}$ to thinking about the real number $\|x-y\|$? We must have applied a function from $\mathbb{R}^{n}$ to $\mathbb{R}$. This is the norm function, which we knew to be continuous from your previous post.



    We have $f_{y}^{-1}\Big((1/j,\infty)\Big)=\{x\in\mathbb{R}^{n}:f_{y}(x)>1/j\}=\{x\in\mathbb{R}^{n}:\|x-y\|>1/j\}=O_{j},$ where the first equality is the definition of pre-image, the second is the definition of $f_{y}(x)$, and the third is the definition of $O_{j}$.

    I hope this post was coherent enough to make sense of and that it helps answer your question of how to "see" the answer.


    Thanks so much for your considerable help, GJA ... most grateful ...

    Am working through your post now ...

    Thanks again,


    Peter

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    #8 Thread Author
    Quote Originally Posted by GJA View Post
    Hi, Peter.



    I will do my best to outline where this methodology/thought process comes from. I apologize in advance if I seem to ramble a bit in places. The first section of the response outlines the thought process from a general point of view. The second section then applies the ideas to your specific question.

    Section 1:

    • The first and most important point bar none is that someone showed/taught me this method (as I am trying to show you now). It's not nearly as difficult to find a hidden treasure when you (basically) know where it's buried. I cannot stress this point enough. Furthermore, I hope it resonates and reassures you that you are doing the right thing by asking questions. I also hope it helps take some of the pressure off your own studies to know that you're seeing the application of an idea, not the invention of it.
    • Once you know to think along these lines, trying to reverse-engineering the reasoning behind them is a great exercise (and is why your question is a good one). Perhaps it would look something like this:

    Q: What is it we want to show?
    A: We want to show this set is open.
    Q: Is there a theorem or definition we know that tells us when a set is open?
    A: Hmm...Aha! We know that, by definition, a continuous function from one space to another is one with the property that the preimage of any open set in the codomain space is open in the domain space.
    Q: How can we actually apply this idea when the time comes?
    A: We can try to recognize the defining characteristic of the set we are interested in (in this case $O_{j}$) as a restriction on the range of a certain function (which we must identify). If we are able to do this, then the set we are interested in is, by definition, the pre-image of the restricted range of the function. Note: This is in some sense akin to computing definite integrals using the Fundamental Theorem of Calculus. You want to try to view the integrand as the derivative of some function (which you must figure out). IF you can find this (so-called) antiderivative, the theorem tells you how to calculate the original integral.

    This method of seeing a new technique, then trying to reverse-engineer the reasoning behind it is one of the best ways to truly internalize new ideas. Creating a sequence of smaller leading questions to get to a final question whose answer you know (i.e., the new method you've been introduced to) can be quite helpful. If you're familiar with the film, think Tom Cruise in the final scene of "A Few Good Men." He knows the answer his witness wants to give, he just needs to ask the right sequence of smaller leading questions before the witness will say the thing that ties the entire case together.

    Section 2:
    • Being primed to think in terms of pre-images of continuous functions, the primary step to solving the problem regarding the openness of $O_{j}$ is converted into a question of whether we can view the defining characteristic of points in the set $O_{j}$ as some sort of restriction on a function (as mentioned above, this is like you knowing the Fundamental Theorem of Calculus already and you just need to know if you can think of the integrand as the derivative of some function or not). The condition defining $O_{j}$ is the set of all $x\in\mathbb{R}^{n}$, such that

    $\|x-y\|>1/j,$

    where $y$ is some fixed element of $\mathbb{R}^{n}$. In words, the inequality is saying that the real number $\|x-y\|$ must be strictly larger than $1/j.$ This leads us to consider the interval $(1/j,\infty).$ Now, how did we go from thinking of $x$ and $y$ as being elements of $\mathbb{R}^{n}$ to thinking about the real number $\|x-y\|$? We must have applied a function from $\mathbb{R}^{n}$ to $\mathbb{R}$. This is the norm function, which we knew to be continuous from your previous post.



    We have $f_{y}^{-1}\Big((1/j,\infty)\Big)=\{x\in\mathbb{R}^{n}:f_{y}(x)>1/j\}=\{x\in\mathbb{R}^{n}:\|x-y\|>1/j\}=O_{j},$ where the first equality is the definition of pre-image, the second is the definition of $f_{y}(x)$, and the third is the definition of $O_{j}$.

    I hope this post was coherent enough to make sense of and that it helps answer your question of how to "see" the answer.

    Hi GJA ... just now staring at your (very clear) explanation and wondering how I missed the answer ....

    Now clear! ... thanks to a very coherent explanation ...

    Thanks!

    Peter

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    #9
    Hi, Peter.

    I am attaching a picture I think might help illuminate the situation. Note that the ball about $p$ should actually be dashed (it's an open ball), though this doesn't affect the comments that follow below.

    Quote Originally Posted by Peter View Post
    You write:

    " ... ... the triangle inequality can be used to show that the ball $\{z:\|z-p\|<\epsilon\}$ and the set $\{x:\|x-y\|\leq 1/j\}$ do not intersect ... "

    Can you help me to demonstrate the above ... ?
    To prove the two sets are disjoint, we must start with an arbitrary point, say $z$, from the ball centered at $p$ and show that its distance to $y$ is greater than $1/j$; i.e., $\|y-z\|>1/j.$ The triangle inequality says

    $\|y-p\|\leq \|z-p\| + \|y-z\|.$

    Now try subtracting $\|z-p\|$ to the other side to obtain a lower bound on $\|y-z\|$ and try to use what we know about $\epsilon$ (from the choice we made in the previous post) to show that this lower bound is strictly larger than $1/j$. Feel free to let me know if this is too cryptic.


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    #10 Thread Author
    Quote Originally Posted by GJA View Post
    Hi, Peter.

    I am attaching a picture I think might help illuminate the situation. Note that the ball about $p$ should actually be dashed (it's an open ball), though this doesn't affect the comments that follow below.



    To prove the two sets are disjoint, we must start with an arbitrary point, say $z$, from the ball centered at $p$ and show that its distance to $y$ is greater than $1/j$; i.e., $\|y-z\|>1/j.$ The triangle inequality says

    $\|y-p\|\leq \|z-p\| + \|y-z\|.$

    Now try subtracting $\|z-p\|$ to the other side to obtain a lower bound on $\|y-z\|$ and try to use what we know about $\epsilon$ (from the choice we made in the previous post) to show that this lower bound is strictly larger than $1/j$. Feel free to let me know if this is too cryptic.



    Hi GJA ... thanks again for your help

    I think that now I have reflected on and understood your points ...

    The proof you asked for goes something like this ... (I hope I'm on the right track!)


    We want to show that the ball $ \displaystyle \{z \ : \ \mid \mid z - p \mid \mid \lt \epsilon \} $

    and

    the set $ \displaystyle \{ x \ : \ \mid \mid x - y \mid \mid \lt \frac{1}{j} \}$

    are disjoint ... ...

    ... in other words, we need to show that

    $ \displaystyle \mid \mid y - z \mid \mid \gt \frac{1}{j}$



    Now $ \displaystyle p \in O_j \Longrightarrow$ can find $ \displaystyle \epsilon$ such that $ \displaystyle 0 \lt \epsilon \lt \mid \mid p - y \mid \mid - \frac{1}{j}$

    $ \displaystyle \Longrightarrow \mid \mid p - y \mid \mid \gt \epsilon + \frac{1}{j}$



    Now ...

    $ \displaystyle \mid \mid y - p \mid \mid \ = \ \mid \mid y - z + z - p \mid \mid \ = \ \mid \mid (z - p) + (y - z) \mid \mid $


    $ \displaystyle \Longrightarrow \mid \mid y - p \mid \mid \ \le \ \mid \mid z - p \mid \mid + \mid \mid y - z \mid \mid$


    $ \displaystyle \Longrightarrow \mid \mid y - z \mid \mid \ \gt \ \mid \mid y - p \mid \mid - \mid \mid z - p \mid \mid$


    $ \displaystyle \Longrightarrow \mid \mid y - z \mid \mid \ \gt \ \epsilon + \frac{1}{j} - \epsilon$


    and so


    $ \displaystyle \mid \mid y - z \mid \mid \ \gt \ \frac{1}{j}$



    Hope that is correct ...


    Thanks again for your guidance and help ...

    Peter
    Last edited by Peter; February 12th, 2018 at 23:11.

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