# Topology questions

#### Poirot

##### Banned
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

##### MHB Oldtimer
Staff member
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.

#### Poirot

##### Banned
Re: Tolpology questions

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

#### Opalg

##### MHB Oldtimer
Staff member
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.

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.

#### Poirot

##### Banned
Re: Tolpology questions

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

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
Staff member
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 $g^{-1}(0)$?

#### Poirot

##### Banned
Re: Tolpology questions

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

BTW I will put the other questions in seperate threads

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#### Opalg

##### MHB Oldtimer
Staff member
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?).

#### Poirot

##### Banned
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.