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Epislon & Delta for Open / Not Open Sets |
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| May21-12, 06:06 PM | #1 |
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Epislon & Delta for Open / Not Open Sets
Hello All,
I am finding the hardest time in understanding how to work δ & ε Open Set Problems? Can someone please explain this approach to me? Thanks in Advance |
| May21-12, 08:34 PM | #2 |
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Hello!
That's kind of vague. Is there a particular problem you are working on? Or is there a Theorem you are trying to understand? Some examples would be easier to respond to. |
| May21-12, 10:36 PM | #3 |
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So the problem given is:
Is the set S = {P=(x,y): |y| > 0} open or not? I want to know how do I tackle or decipher this?...I have spent a good 2 days researching and reviewing the text book example and I am simply not getting it. |
| May21-12, 10:44 PM | #4 |
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Epislon & Delta for Open / Not Open Sets |
| May21-12, 10:52 PM | #5 |
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Any arbitrary point P within a set must be < than the δ or radius in order to satisfy as an "Open Set". Was this what you are asking? I am simply confused and lost on how one would write a proof and hence I need a better explanation on understanding this types of "Prove" |
| May22-12, 05:09 AM | #6 |
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I suggest going back and re-reading (and understand?) the formal definition. What you have said above is not quite accurate. The structure of the proof (ie. what you have to do) comes really from the definition itself.
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| May22-12, 11:01 AM | #7 |
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My guess about the definition in your text materials is that it says something equivalent to: A set S is open if and only if for each point x in S, there exists an open interval O of radius [itex] \delta > 0 [/itex] such that O contains x and O is a subset of S. (The definition in you book might want x to be the center of O.) The way you start such a proof is to say "let x be a point in the set S = {w: |w| > 0}". Then you show you can find an open interval O that meets these condition. You can divide the proof into 2 cases, if you wish. For example, if you take the case that x < 0 then you can say "Consider the case that x < 0. Since x < 0, -x/2 > 0. Let [itex] \delta = -x/2 [/itex]." Then you must show that an open interval O with center x and radius [itex] \delta [/itex] is a subset of S. Proofs involving showing one set is a subset of another often are done by showing that if we pick an arbitrary point in the first set then it must be in the second set. So you may began this part of the proof by saying "Let p be a point in 0 = {y:| y - x | < [itex] \delta [/itex] }." Then you prove p is in S by showing |p| > 0. If it's any consolation, writing the proof in such detail is not easy. You should consider whether your textbook has proved any theorems that could be used. Without seeing your text, I don't know what it has proven. For example, has the textbook proven that a half open interval such as {q: q > C} is an open set? |
| May22-12, 04:42 PM | #8 |
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Equivalent to what Stephen Tashi said, a set, A, is open if and only if, for any point p in A, there exist [itex]\delta> 0[/itex] such that, for any point q, if, [itex]d(p, q)< \delta[/itex] the q is in A also.
Many textbooks start by defining an "interior point" of a set, A, as any point, p, for which the above is true, then define A to be "open" if all of its points are interior points. All this, by the way, is only true for metric topologies in which d(p, q), the distance between points p and q, is defined. There are more general definitions. |
| May22-12, 05:39 PM | #9 |
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Ok guys, my apologies. I did more research on this problem set and this is exactly what is given.
Is the set S = {P=(x,y): |y| > 0} open or not? {P=(x,y): |y| > 0} = {P=(x,y): (y > 0) or (y <0)} = {P=(x,y):y ≠ 0} I need to show whether this is a open set or not? Hope this iis more clearer. |
| May22-12, 05:55 PM | #10 |
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firstly: THANK YOU all for taking the time and trying to point me in the right direction.
Now, I am sitting here pondering on this problems "Open Set" problems. I believe the question I should be asking is whether someone can explain to me Open Sets using δ & ε? with examples and solution Than maybe I can wrap my head around this. TIA |
| May22-12, 07:34 PM | #11 |
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Now if x is a point in S, can you draw a little circle around it that's entirely within S? |
| May22-12, 07:52 PM | #12 |
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| May22-12, 07:54 PM | #13 |
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Can you turn this into a proof? Is that where you're stuck? |
| May22-12, 09:54 PM | #14 |
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Think about it this way, for intuition, which you can then formalize to create your "proof".
You want to make sure that for all points in that given set S, you can create a "disk", or "neighborhood" [ with radius δ > 0 ], around each one of them, in such a way that all the points inside this disk is contained in S. That is essentially what it means to be an "open" set. So, for this kind of proof, it suffices to show that there exist such a "neighborhood" for an arbitrary point P in S. Pick such a point. Now, ask yourself: Where can you run into trouble when creating the "neighborhood"? Is it possible for you to circumvent that problem for any choice of point P? In other words, given a point (x,y), what must delta be so that you don't run into the problem? |
| May22-12, 10:47 PM | #15 |
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I believe that "let P be any point in S such that (P-P0)/2 < δ"..again for the life of me I can't express in or formulate it into a statement. |
| May22-12, 10:52 PM | #16 |
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What is P0? You need to make sure every detail in your proof is accounted for.
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| May22-12, 11:00 PM | #17 |
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