The Thread Ask Mortimer a Question

[miketx]

My head explodes that is too much for my very low IQ.

Very well. We will transport you Fomalhaut 3, where you can labor in the with the Denarian birthing tubes.

 

[gtopa1]

What is the meaning of life?

42.

Greg

 

[Mortimer]

Explain to us why Urysohn's metrization theorem states that every Hausdorff second-countable regular space is metrizable?

I really do not know what that means, I didnt ignored you but that is too high for me.

 

[gtopa1]

Ok a question.

If your jump drive is over heating and you drop out of subspace in sector 9-843, do you head for the nearby star to use the radiation from it to recharge your systems or will you wait on your jump drive to cool, or do you take a chance your jump signature hasn't been detected and go ahead and release all counter measures hoping the Anatari won't detect you?

I'd call "The Doctor".

Greg

 

[miketx]

42.

Greg

I hope you took into account inflation.

 

[gtopa1]

Explain to us why Urysohn's metrization theorem states that every Hausdorff second-countable regular space is metrizable?

Is that the question???

First proof Proof of Theorem 2.1. As we said, this proof will proceed by constructing a continuous injection F : X → R N prod that is a homeomorphism onto its range. Let { fn : n ∈ N } be a collection of functions as in Lemma 2.3. We could of course use the specific set of functions defined in the proof of the Lemma, but from this point onwards we do not need to be that specific. Define F : X → R N by F(x) = (f1(x), f2(x), f3(x), . . .). This function is continuous since each of its component functions is continuous. It is also injective since for any x 6= y ∈ X, we can find an open set U containing x but not y, and therefore we can find an n ∈ N such that fn(x) > 0 and fn(y) = 0, and therefore F(x) 6= F(y). Let Y = F(X) ⊆ R N be the range of F. We already know that Y is metrizable (with its subspace topology), since R N prod is metrizable, and we just established that F is continuous and injective. All that remains to show is that F is open.

So fix an open set U ⊆ X. We show that F(U) is open in Y . We will do this in the usual way, by fixing an arbitrary point b ∈ F(U) and finding an open subset V of Y such that b ∈ V ⊆ F(U). Let a = F −1 (b) ∈ U (the preimage is a single point since F is injective). By the lemma, there is some N ∈ N such that fN (a) > 0 and fN (x) = 0 for all x ∈ X \ U. We define our open set V ⊆ Y by: V := Y ∩ π −1 N ((0, ∞)). This set is open in the subspace topology on Y since π −1 N ((0, ∞)) is a basic open subset of R N prod. Note that b ∈ V since πN (b) = πN (F(a)) = fN (a) > 0. The last thing to show is that V ⊆ F(U). To show this, fix an arbitrary y ∈ V . Since y ∈ Y , we can define x := F −1 (y). Since y ∈ π −1 N ((0, ∞)) we have that 0 < πN (y) = πN (F(x)) = fN (x). But we know that fN is zero for all points outside of U, and therefore it must be the case that x ∈ U, and in turn that y = F(x) ∈ F(U), as required. This shows that F : X → Y is a homeomorphism, and therefore that X is metrizable. Note that we actually embedded X as a subspace of [0, 1]N ⊆ R N. It should not be so surprising that this is possible, given that we know any metrizable space can be generated by a bounded metric. The space (0, 1)N is homeomorphic to R N, and so if anything embedding into [0, 1]N gives us more room to work with. 4 Second proof Before we begin, note that we may assume the functions { fn : n ∈ N } provided by Lemma 2.3 actually map into [0, 1 n ] rather than [0, 1], since we can simply replace fn by 1 n fn if necessary. Proof of Theorem 2.1. In this proof, we embed X into the “Hilbert cube” H := Q n∈N [0, 1 n ] with the topology generated by the uniform metric. To be clear, let du be the metric defined on H by du(x, y) = sup { |xn − yn| : n ∈ N } , where x = (x1, x2, x3, . . .) as usual, and similarly for y. Throughout this proof, we will assume H has the metric topology generated by du (or in other words the subspace topology inherited from R N unif). Recall also that the uniform topology refines the product topology. We define F : X → H in the same way as before: F(x) = (f1(x), f2(x), f3(x), . . .), c 2018– Ivan Khatchatourian 3 15. Urysohn’s metrization theorem 15.4. Second proof (where remember that we are assuming fn maps into [0, 1 n ] for each n). Our previous argument also shows that F is injective, and F is still open since the uniform topology refines the product topology. It remains only to show that F is continuous, which is no longer obvious since we are not using the product topology. We prove that F is continuous using the “continuity at a point” characterization of continuity. So fix a ∈ X, and fix > 0. We want to find an open set U containing a such that x ∈ U =⇒ du(F(a), F(x)) < , or in other words that F(U) ⊆ B(F(a)). This argument proceeds similarly to the proof that R N prod is metrizable, by splitting things up to a finite part we can explicitly take care of, and an infinite tail which is so small that it doesn’t matter. Fix an N ∈ N so large that 1 N < 2 . Then for each n ≤ N, since fn is continuous we can find an open set Un containing a such that |fn(a) − fn(x)| < 2 for all x ∈ Un. Then U := U1 ∩ U2 ∩ · · · ∩ UN is our desired open set. Indeed, if x ∈ U, then by construction |fn(x) − fn(a)| < 2 for all n ≤ N. On the other hand if n > N, then |fn(x) − fn(a)| ≤ 1 n < 1 N < 2 since fn maps into [0, 1 n ]. In total, we have that |fn(x) − fn(a)| < 2 for all n ∈ N. Therefore du(F(x), F(a)) = sup { |fn(x) − fn(a)| : n ∈ N } ≤ 2 < , as required

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In short I have no idea if that answers your question but I would like to know why you expected an answer??

Am I right to be amused by your question?

Greg

 

[gtopa1]

I hope you took into account inflation.

Dammit no!!!

Make that ...oh no. Supercomputer is down.

Sorry about that, folk.

Greg

 

[miketx]

Dammit no!!!

Make that ...oh no. Supercomputer is down.

Sorry about that, folk.

Greg

On my transport we use an AI. Sombish knows it all!

 

[gtopa1]

On my transport we an AI. Sombish knows it all!

Ask them to redo the calculations. The Master has the starting Algorithm.

Greg

 

[luiza]

Only a few sentences.

That is what you need . Just one long sentence .

 

[Tunafish]

I dont know you tell me?

I asked you

 

[JGalt]

Ok a question.

If your jump drive is over heating and you drop out of subspace in sector 9-843, do you head for the nearby star to use the radiation from it to recharge your systems or will you wait on your jump drive to cool, or do you take a chance your jump signature hasn't been detected and go ahead and release all counter measures hoping the Anatari won't detect you?

Brilliant question! I was just gonna ask him if all Austrians were Sofa King We Todd Did.

 

[miketx]

Ask them to redo the calculations. The Master has the starting Algorithm.

Greg

Comes out the same. Still says Mortimer is weird.

 

[miketx]

Is that the question???

First proof Proof of Theorem 2.1. As we said, this proof will proceed by constructing a continuous injection F : X → R N prod that is a homeomorphism onto its range. Let { fn : n ∈ N } be a collection of functions as in Lemma 2.3. We could of course use the specific set of functions defined in the proof of the Lemma, but from this point onwards we do not need to be that specific. Define F : X → R N by F(x) = (f1(x), f2(x), f3(x), . . .). This function is continuous since each of its component functions is continuous. It is also injective since for any x 6= y ∈ X, we can find an open set U containing x but not y, and therefore we can find an n ∈ N such that fn(x) > 0 and fn(y) = 0, and therefore F(x) 6= F(y). Let Y = F(X) ⊆ R N be the range of F. We already know that Y is metrizable (with its subspace topology), since R N prod is metrizable, and we just established that F is continuous and injective. All that remains to show is that F is open.

So fix an open set U ⊆ X. We show that F(U) is open in Y . We will do this in the usual way, by fixing an arbitrary point b ∈ F(U) and finding an open subset V of Y such that b ∈ V ⊆ F(U). Let a = F −1 (b) ∈ U (the preimage is a single point since F is injective). By the lemma, there is some N ∈ N such that fN (a) > 0 and fN (x) = 0 for all x ∈ X \ U. We define our open set V ⊆ Y by: V := Y ∩ π −1 N ((0, ∞)). This set is open in the subspace topology on Y since π −1 N ((0, ∞)) is a basic open subset of R N prod. Note that b ∈ V since πN (b) = πN (F(a)) = fN (a) > 0. The last thing to show is that V ⊆ F(U). To show this, fix an arbitrary y ∈ V . Since y ∈ Y , we can define x := F −1 (y). Since y ∈ π −1 N ((0, ∞)) we have that 0 < πN (y) = πN (F(x)) = fN (x). But we know that fN is zero for all points outside of U, and therefore it must be the case that x ∈ U, and in turn that y = F(x) ∈ F(U), as required. This shows that F : X → Y is a homeomorphism, and therefore that X is metrizable. Note that we actually embedded X as a subspace of [0, 1]N ⊆ R N. It should not be so surprising that this is possible, given that we know any metrizable space can be generated by a bounded metric. The space (0, 1)N is homeomorphic to R N, and so if anything embedding into [0, 1]N gives us more room to work with. 4 Second proof Before we begin, note that we may assume the functions { fn : n ∈ N } provided by Lemma 2.3 actually map into [0, 1 n ] rather than [0, 1], since we can simply replace fn by 1 n fn if necessary. Proof of Theorem 2.1. In this proof, we embed X into the “Hilbert cube” H := Q n∈N [0, 1 n ] with the topology generated by the uniform metric. To be clear, let du be the metric defined on H by du(x, y) = sup { |xn − yn| : n ∈ N } , where x = (x1, x2, x3, . . .) as usual, and similarly for y. Throughout this proof, we will assume H has the metric topology generated by du (or in other words the subspace topology inherited from R N unif). Recall also that the uniform topology refines the product topology. We define F : X → H in the same way as before: F(x) = (f1(x), f2(x), f3(x), . . .), c 2018– Ivan Khatchatourian 3 15. Urysohn’s metrization theorem 15.4. Second proof (where remember that we are assuming fn maps into [0, 1 n ] for each n). Our previous argument also shows that F is injective, and F is still open since the uniform topology refines the product topology. It remains only to show that F is continuous, which is no longer obvious since we are not using the product topology. We prove that F is continuous using the “continuity at a point” characterization of continuity. So fix a ∈ X, and fix > 0. We want to find an open set U containing a such that x ∈ U =⇒ du(F(a), F(x)) < , or in other words that F(U) ⊆ B(F(a)). This argument proceeds similarly to the proof that R N prod is metrizable, by splitting things up to a finite part we can explicitly take care of, and an infinite tail which is so small that it doesn’t matter. Fix an N ∈ N so large that 1 N < 2 . Then for each n ≤ N, since fn is continuous we can find an open set Un containing a such that |fn(a) − fn(x)| < 2 for all x ∈ Un. Then U := U1 ∩ U2 ∩ · · · ∩ UN is our desired open set. Indeed, if x ∈ U, then by construction |fn(x) − fn(a)| < 2 for all n ≤ N. On the other hand if n > N, then |fn(x) − fn(a)| ≤ 1 n < 1 N < 2 since fn maps into [0, 1 n ]. In total, we have that |fn(x) − fn(a)| < 2 for all n ∈ N. Therefore du(F(x), F(a)) = sup { |fn(x) − fn(a)| : n ∈ N } ≤ 2 < , as required

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In short I have no idea if that answers your question but I would like to know why you expected an answer??

Am I right to be amused by your question?

Greg

Dude, everybody knows that shit. Like duh.

 

[Remodeling Maidiac]

You ask Mortimer a question, and I answer. Thanks.

What do you do for a living?

 

[Mortimer]

What do you do for a living?

I paint the sky blue.

 

[miketx]

I paint the sky blue.

Another words you live in your moms basement and mooch.

 

[JGalt]

I paint the sky blue.

Hmmmm. Hitler was quite the painter too.

Stay out of beer halls, son.

 

[Natural Citizen]

You ask Mortimer a question, and I answer. Thanks.

Do fish drink water?

 

[miketx]

Do fish drink water?

I dunno but they swims around in it with their big mouths open so I guess dey haf to.