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Topology questions

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Poirot

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Feb 15, 2012
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1) Endow Z with the toplogy T consisting of the empty set and all subsets containing 0.

Let g(Z,T)->(Z,T) be continouous and bijective. Prove that g is a homeomorphism.

Clearly I have to prove $g^{-1}$
is continuous but I can't see how to.

2) let f map polynomials to polynomials. Prove that if f(p(x))=1+1/2$\int_{0}\,^{x} p(t)\,dt$, the f is a contraction mapping with no fixed points

3)
Let B be a subset of X. For each B define the topolgy τ to consist of the subsets U of X such that U∩B is empty, plus the empty set. Let A be an infinite subset of X. Show that A is compact in X if and only if A ∩ B is not equal to the ∅.
 

Opalg

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Feb 7, 2012
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Re: Tolpology questions

1) Endow Z with the toplogy T consisting of the empty set and all subsets containing 0.

Let g(Z,T)->(Z,T) be continouous and bijective. Prove that g is a homeomorphism.

Clearly I have to prove $g^{-1}$ is continuous but I can't see how to.
Notice that the definition of the topology T is closely tied to sets containing $0$. You are told that $g$ is continuous with respect to T. What does that tell you about $g(0)$? What can you then deduce about $g^{-1}(0)$? Use that information to show that $g^{-1}$ is continuous.

2) let f map polynomials to polynomials. Prove that if f(p(x))=1+1/2$\int_{0}\,^{x} p(t)\,dt$, the f is a contraction mapping with no fixed points
In order to define a contraction mapping, you need to have a metric on the space. You have not told us what the metric is in this case.
 
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Poirot

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Feb 15, 2012
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Re: Tolpology questions

let's concentrate on the first for now. Is it intermediatle value theorem?
 

Opalg

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Feb 7, 2012
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Re: Tolpology questions

let's concentrate on the first for now. Is it intermediatle value theorem?
No, it could hardly be further away from anything like the intermediate value theorem. (Shake)

The mapping $g$ goes from the set $Z$ of integers to itself. So it is very discrete in nature, and certainly cannot take "intermediate" values. You are told that $g$ is continuous, but that has nothing to do with the usual idea of continuity. It means continuity with respect to the metric T.

The definition of continuity for a mapping $g$ on a metric space says that if $U$ is an open set then $g^{-1}(U)$ is open. Here, you need to use the definition of openness for the topology T, namely that a subset of $Z$ is open if it contains 0.
 
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Poirot

Banned
Feb 15, 2012
250
Re: Tolpology questions

No, it could hardly be further away from anything like the intermediate value theorem. (Shake)

The mapping $g$ goes from the set $Z$ of integers to itself. So it is very discrete in nature, and certainly cannot take "intermediate" values. You are told that $g$ is continuous, but that has nothing to do with the usual idea of continuity. It means continuity with respect to the metric T.

The definition of continuity for a mapping $g$ on a metric space says that if $U$ is an open set then $g^{-1}(U)$ is open. Here, you need to use the definition of openness for the topology T, namely that a subset of $Z$ is open if it contains 0.
I need to show for every open set U in (z,T), g(U) is open n (Z,T). If U is empty, it is trivial. Otherwise U contains 0. But I don't know anything about g(U)
 

Opalg

MHB Oldtimer
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Feb 7, 2012
2,725
Re: Tolpology questions

I need to show for every open set U in (z,T), g(U) is open n (Z,T). If U is empty, it is trivial. Otherwise U contains 0. But I don't know anything about g(U)
Start by looking at what you do know. You know that $g$ is continuous. This means that $g^{-1}(U)$ is open whenever $U$ is open. In other words, $g^{-1}(U)$ contains 0 whenever $U$ contains 0. What does that tell you about $g^{-1}(0)$?
 
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Poirot

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Feb 15, 2012
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Re: Tolpology questions

$g^{-1}(0)$ is the union of all open sets?

BTW I will put the other questions in seperate threads
 
Last edited:

Opalg

MHB Oldtimer
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Feb 7, 2012
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Re: Tolpology questions

$g^{-1}(0)$ is the union of all open sets?
More like the intersection of all open sets (excluding the empty set).

We got as far as saying that $g^{-1}(U)$ contains 0 whenever $U$ contains 0. In particular, the set $U=\{0\}$ contains $0$. Therefore $g^{-1}(\{0\})$ contains $0$. But $g$ is bijective, so $g^{-1}(0)$ consists of a single point. That single point must be (what?).
 
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Poirot

Banned
Feb 15, 2012
250
Re: Tolpology questions

0. So therefore g(0)=0. 0 is in all non-empty open sets, so 0 must be in their image. Q.E.D.