Continuous Mappings

  • Thread starter Oxymoron
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  • #1
Oxymoron
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I would like someone to check my working please. Here is the question:

Is the mapping

[tex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/tex]

continuous?

My solution:

A mapping [itex]f:X\rightarrow Y[/itex] is continuous at a point [itex]x \in X[/itex] if for every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that

[tex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/tex]

A function is continuous if it is continuous at every point in its domain.

This is the definition of continuity that I will be using. Note that I made this choice because the domain of the mapping lies in the rationals with respect to the p-adic metric. Further, [itex]\mathbb{Q}_p[/itex] and [itex]\mathbb{R}[/itex] are both metric spaces - ie there is a very good notion of distance between two points in both sets. This idea is crucial to my solution.

Loosely, the definition above says that all points [itex]f(y) \in Y[/itex] near [itex]f(x) \in Y[/itex] have their preimages [itex]y \in X[/itex] near [itex]x \in X[/itex].

In the question at hand, the function [itex]f[/itex] maps points in the metric space [itex]\mathbb{Q}_p[/itex] to points in [itex]\mathbb{R}[/itex] - where [itex]\mathbb{R}[/itex] is the metric space w.r.t to usual Euclidean metric.

For the function to be continuous the following must be satisfied: The point [itex]f(x) \in \mathbb{R}[/itex] near [itex]f(x) \in \mathbb{R}[/itex] must imply that [itex]y \in \mathbb{Q}_p[/itex] is near [itex]x \in \mathbb{Q}_p[/itex].

Technically speaking, given an epsilon > 0 there must exist a delta > 0 such that [itex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/itex].

But [itex]d(x,y) := |x-y|_p[/itex]. That is, distance between two points with respect to the p-adic metric gets larger and larger as [itex]x[/itex] gets closer and closer to [itex]y[/itex] with respect to the Euclidean metric. So I can not always guarantee the existance of a positive, non-zero [itex]\delta[/itex] for any given [itex]\epsilon[/itex]. Therefore the mapping [itex]f[/itex] is NOT continuous.

To make this clearer, the distance between two points [itex]x, y \in \mathbb{Q}_p[/itex] actually gets larger as [itex]x[/itex] and [itex]y[/itex] become closer with respect to the Euclidean metric. So the preimages of 'close' points in [itex]\mathbb{R}[/itex] are actually very far apart in [itex]\mathbb{Q}_p[/itex]. This is the exact opposite of what it means for a function to be continuous.

How does this sound?
 
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Answers and Replies

  • #2
shmoe
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It looks like [tex]f[/tex] is the inclusion map from the rationals to the reals, but considering the rationals with the p-adic metric. If so writing [tex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/tex] doesn't make sense as it's not a function on all of [tex]\mathbb{Q}_p[/tex].

To show it's not continuous at a point x, you have to prove that you can find y's arbitrarily close in the Euclidean metric that are arbitrarily far away in the p-adic metric (it's sufficient to find y's farther than some convenient fixed bound, 1 will work here). You've essentially just declared this was true, but offered no proof. Can you supply details?
 
  • #3
Oxymoron
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Oh dear, I forgot to mention that

[tex]f(x) = x[/tex].
 
  • #4
Oxymoron
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May I ask if this is actually a continuous function?
 
  • #5
Oxymoron
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To show it's not continuous at a point x, you have to prove that you can find y's arbitrarily close in the Euclidean metric that are arbitrarily far away in the p-adic metric (it's sufficient to find y's farther than some convenient fixed bound, 1 will work here). You've essentially just declared this was true, but offered no proof. Can you supply details?

I actually considered that this was obvious. But you are right, proving this is much more concrete.

We need to show that given an [itex]\epsilon > 0[/itex] we need to find a [itex]\delta > 0[/itex] so that whenever [itex]d(x,y) < \delta[/itex] then [itex]d(1(x),1(y)) < \epsilon[/itex]. However, since [itex]1(x) = x[/itex] by definition of the function, it follows that [itex]d(1(x),1(y)) = d(x,y) = |x-y|_p[/itex]. Therefore we cannot necessarily find a [itex]\delta[/itex] which is strictly less that [itex]\epsilon[/itex].
 
  • #6
shmoe
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Oxymoron said:
I actually considered that this was obvious. But you are right, proving this is much more concrete.

We need to show that given an [itex]\epsilon > 0[/itex] we need to find a [itex]\delta > 0[/itex] so that whenever [itex]d(x,y) < \delta[/itex] then [itex]d(1(x),1(y)) < \epsilon[/itex]. However, since [itex]1(x) = x[/itex] by definition of the function, it follows that [itex]d(1(x),1(y)) = d(x,y) = |x-y|_p[/itex]. Therefore we cannot necessarily find a [itex]\delta[/itex] which is strictly less that [itex]\epsilon[/itex].

It's probably best to not use [tex]d(\cdot,\cdot)[/tex] to represent two different metrics, so I'm going to switch to [tex]|\cdot|_p[/tex] for the p-adic norm and [tex]|\cdot|[/tex] for the usual euclidean norm.

You have the same problem as before. "cannot necessarily find..." isn't a proof. f is continuous at x if for every [tex]\epsilon>0[/tex] we can find a [tex]\delta>0[/tex] such that for all y with [tex]|x-y|_p<\delta[/tex] we have [tex]|x-y|=|f(x)-f(y)|<\epsilon[/tex].

To prove f is NOT continuous at x we can find an [tex]\epsilon>0[/tex] that has no [tex]\delta>0[/tex] that will satisfy this. Showing it fails for any specific [tex]\epsilon[/tex] will be enough, so let's take [tex]\epsilon=1[/tex]. Now suppose [tex]\delta>0[/tex] satisfies the definition of continuity for [tex]\epsilon=1[/tex]. Can you find a y where [tex]|x-y|_p<\delta[/tex] but we have [tex]|x-y|>1[/tex]?
 
  • #7
Oxymoron
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Good point Shmoe!

If I take y = 2 then (2 must be prime)

[tex]|0-2|_p = \frac{1}{2} < \delta[/tex]

But

[tex]|0-2| = 2 > 1[/tex]

Is this ok?
 
  • #8
Hurkyl
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Oh dear, I forgot to mention that

[tex]f(x) = x[/tex]
Is this even a well-defined function? x is a p-adic number... in what sense are you embedding the p-adics into the reals?


Anyways, it's very easy to exhibit a sequence that converges in the p-adics and not in the reals. For example, consider the 3-adic sequence:

2, 22, 222, 2222, 22222, ...
which converges to the 3-adic number ...222 (= -1)

But this sequence clearly does not converge in the reals.


On the other hand...


I haven't worked it out, but it seems to me that if I define a function from Qp &rarr; R that maps a p-adic number x to its lexicographical reverse as a real number written in radix p, then this function is a continuous surjection. (not a bijection)

For example, in the 5-adics,

f(...41302.134) = 431.20314...(base 5)

This is in line with my intuition that the p-adics and the reals are opposites.
 
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  • #9
shmoe
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Oxymoron: You've done a very specific case, when x=0, p=2 and delta>1/2. Let's carry on with this, leaving x=0, and p=2.

Find me a y where [tex]|0-y|_p<1/2[/tex] but [tex]|0-y|>1[/tex].

Find me a y where [tex]|0-y|_p<1/4[/tex] but [tex]|0-y|>1[/tex].

Find me a y where [tex]|0-y|_p<\delta[/tex] but [tex]|0-y|>1[/tex]. y will depend on [tex]\delta[/tex]


I'm also a little concerned with what you wrote, you know that the p-adic norm [tex]|\cdot |_p[/tex] depends on a specific prime p right? That is, [tex]|\cdot |_2[/tex], [tex]|\cdot |_3[/tex], [tex]|\cdot |_5[/tex] etc. all give different norms.
 
  • #10
Oxymoron
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Is this even a well-defined function? x is a p-adic number... in what sense are you embedding the p-adics into the reals?

The question simply asks to determine whether or not [itex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/itex] with f(x) = x is a continuous function.

I haven't worked it out, but it seems to me that if I define a function from Qp → R that maps a p-adic number x to its lexicographical reverse as a real number written in radix p, then this function is a continuous surjection. (not a bijection)

Clever. I haven't seen or heard anything like that!

This is in line with my intuition that the p-adics and the reals are opposites.

Interesting... how sure are you of this claim (along with the function being surjective if it maps p-adic numbers to their real lexicographic reverse). Im gonna look into this a bit more!

I'm also a little concerned with what you wrote, you know that the p-adic norm depends on a specific prime p right? That is, , , etc. all give different norms.

Yep. thats right.

Ok, I see where you are going now Shmoe.

I have indeed chosen a very specific case where I can actually picked a certain prime (2 in this case). I could also have chosen some other prime.

[tex]|0-y|_2 < 1/2 [/tex]

Take [itex]y = 4[/itex] then

[tex]|0 - 4|_2 = \frac{1}{2^2} = 2^{-2} < 1/2[/tex]

But

[tex]|0-4| = 4 > 1[/tex]

Similarly for your second equation I could choose y = 5.
 
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  • #11
Hurkyl
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The question simply asks to determine whether or not [itex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/itex] with f(x) = x is a continuous function.

Then I would have to say no, because it's not a function at all. It would make sense to ask something like:

Is [itex]f:\mathbb{Q}_p \cap \mathbb{Q} \rightarrow \mathbb{R} : x \rightarrow x[/itex] a continuous function?

Because here, the domain is a subset of the range. But [itex]\mathbb{Q}_p[/itex] contains horrible things (it's uncountable, just like the reals), and given one, I can't see any "right" way for viewing it as an element of [itex]\mathbb{R}[/itex].


A an interesting note, the metric closure of the algebraic closure of [itex]\mathbb{Q}_p[/itex] (called [itex]\Omega_p[/itex] is apparently field isomorphic to [itex]\mathbb{C}[/itex], but Wikipedia doesn't say whether there is an isomorphism as a topological field. (It does say that the existance of the field isomorphism requires the axiom of choice, so it's safe to say that there isn't a nice, canonical isomorphism of the two!)


Interesting... how sure are you of this claim (along with the function being surjective if it maps p-adic numbers to their real lexicographic reverse).

I'm optimistic. If it's not actually continuous, I'm pretty sure it's "mostly" continuous, in some intuitive sense.

It should be obvious that the function is surjective, since the p-adics are isomorphic the set of all left-infinite radix p numbers, and the reals are isomorphic the set of all right-infinite radix p numbers (modulo some relations, such as 1 = 0.zzz... where z = p - 1). :tongue2:
 
  • #12
Oxymoron
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I spoke to my topology professor today and he said that this function is not continuous - but he did not talk too much about it (since it is an assignment problem). But he approved of what Shmoe and I have been discussing. I think it is safe to say that this function is not continuous.
 
  • #13
Oxymoron
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I have another function I have been looking at - one which I think IS continuous.

[tex]f:\{a,b\} \rightarrow \mathbb{N}[/tex]

with [itex]g(a) = 1[/itex], [itex]g(b) = 2[/itex], and [itex]\{a,b\}[/itex] has the topology [itex]\{\oslash, \{a,b\}, \{a\}\}[/itex].


I am fairly certain that this function is continuous using the following definition of a continuous function:

For every open subset [itex]U[/itex] of [itex]Y[/itex] the inverse image [itex]f^{-1}(U)[/itex] is an open subset of [itex]X[/itex].


I want to use that fact the the topology contains only open sets, which are open subsets of [itex]\{a,b\}[/itex]. So in my opinion it is obvious that the sets in [itex]\{a,b\}[/itex] are open, but I want this for any choice of [itex]U[/itex] in [itex]\mathbb{N}[/itex].

But then I had another thought.

[itex]g[/itex] maps [itex]a \in \{a,b\}[/itex] to [itex]1 \in \mathbb{N}[/itex], and [itex]g[/itex] also maps [itex]b \in \{a,b\}[/itex] to [itex]2 \in \mathbb{N}[/itex]. Therefore [itex]g^{-1}(1)[/itex] maps to [itex]a[/itex] which is open with respect to the topology there. But [itex]g^{-1}(2)[/itex] maps to [itex]b[/itex] which is not open with respect to the topology. So the function is not continuous.

Now I am confused...
 
  • #14
shmoe
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Oxymoron said:
The question simply asks to determine whether or not [itex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/itex] with f(x) = x is a continuous function.

It almost surely says more than this, something about restricting the function to the rationals? This is what I was pointing out about your notation in post #2 and what Hurkyl has been saying about it not being a function.

Oxymoron said:
Similarly for your second equation I could choose y = 5.

I had meant to keep p fixed at 2, so if you're still working on this

Find me a y where [tex]|0-y|_2<1/4[/tex] but [tex]|0-y|>1[/tex].

Find me a y where [tex]|0-y|_2<\delta[/tex] but [tex]|0-y|>1[/tex]. y will depend on [tex]\delta[/tex]


What's confusing you about your second question? It's not the discrete topology on {a,b}, you have to look at the topology to determine if {b} is open or not.
 
  • #15
Oxymoron
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b is not open in the topology. So the inverse image of 2 is not mapped to an open set. Hence the function is not continuous.
 
  • #16
Oxymoron
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I had meant to keep p fixed at 2, so if you're still working on this...
.

Well if I choose [itex]y = 8[/itex] then

[tex]|0-8|_p = |2^3|_p = 1/8 < 1/4[/tex]

But

[tex]|0-8| = 8 > 1[/tex]

So given a [itex]\delta[/itex] I can find a [itex]y[/itex]:

[tex]y = \frac{2}{\delta}[/tex]

such that

[tex]|0-y|_p < \delta[/tex] but [tex]|0-y|> 1[/tex]
 
  • #17
shmoe
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For your y given a general delta, are there any conditions on delta? Does it work if delta=1/3 for example?
 
  • #18
Oxymoron
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Perhaps I should have written

[tex]y = \frac{p}{\delta}[/tex]

Because I wrote [itex]y = 2/\delta[/itex] since [itex]p = 2[/itex].
 
  • #19
Oxymoron
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Delta must be expressible as [itex]\delta = 1/p^n[/itex] for some [itex]n\in\mathbb{N}[/itex]
 
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  • #20
matt grime
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Oxymoron said:
I have another function I have been looking at - one which I think IS continuous.

[tex]f:\{a,b\} \rightarrow \mathbb{N}[/tex]

with [itex]g(a) = 1[/itex], [itex]g(b) = 2[/itex], and [itex]\{a,b\}[/itex] has the topology [itex]\{\oslash, \{a,b\}, \{a\}\}[/itex].


I am fairly certain that this function is continuous using the following definition of a continuous function:

For every open subset [itex]U[/itex] of [itex]Y[/itex] the inverse image [itex]f^{-1}(U)[/itex] is an open subset of [itex]X[/itex].


this only makes sense if you assign a topology to N, since f is continuous iff the inverse image of an OPEN set is open. How do youknow that {2} is open in N?

I want to use that fact the the topology contains only open sets, which are open subsets of [itex]\{a,b\}[/itex]. So in my opinion it is obvious that the sets in [itex]\{a,b\}[/itex] are open, but I want this for any choice of [itex]U[/itex] in [itex]\mathbb{N}[/itex].

But then I had another thought.

[itex]g[/itex] maps [itex]a \in \{a,b\}[/itex] to [itex]1 \in \mathbb{N}[/itex], and [itex]g[/itex] also maps [itex]b \in \{a,b\}[/itex] to [itex]2 \in \mathbb{N}[/itex]. Therefore [itex]g^{-1}(1)[/itex] maps to [itex]a[/itex] which is open with respect to the topology there. But [itex]g^{-1}(2)[/itex] maps to [itex]b[/itex] which is not open with respect to the topology. So the function is not continuous.

Now I am confused...
 
  • #21
Oxymoron
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Does the inclusion of the last quote without a comment indicate that my reasoning there is correct?
 
  • #22
matt grime
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No, my comment applies to both patrs I quote back at you.


If you want to determine if a fucntion is continuous from X to Y you need to specify the topologies of both X adn Y.

If, in this example, N has the discreate topology then you are coorect that since the inverse image of {2} (an open set in the discrete topology) is {b} and {b} is not an open set in the given topology on {a,b} then the map is not continuous.

As you have not stated the topology on N you cannot come to any conclusion at all.
 
  • #23
Oxymoron
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But the question has not provided me with a topology for [itex]\mathbb{N}[/itex] at all - only a topology for [itex]\{a,b\}[/itex]. Here is the exact question:

Determine if the mapping

[tex]f:\{a,b\} \rightarrow \mathbb{N}[/tex]

with [itex]f(a) = 1,\,f(b) = 2[/itex], and [itex]\{a,b\}[/itex] has the topology [itex]\{\oslash, \{a\},\{a,b,\}\}[/itex].

It does not mention anything about a topology on [itex]\mathbb{N}[/itex].
 
  • #24
Oxymoron
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Thanks Matt for the insight. I am going to assume that [itex]\mathbb{N}[/itex] has the discrete topology and that the function is not continuous for now.

But I want to highlight another mapping:

[tex]f:\mathbb{N}\rightarrow \mathbb{N}[/tex]

with [itex]f(x) = x^2[/itex], and now [itex]\mathbb{N}[/itex] has the finite complement topology.
 
  • #25
matt grime
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We can assume it is the discrete topology, I would imagine. Possibly at some point in the book it mentions something like: all discrete sets with have the discrete topology unless specified otherwise. Then {2} is open and the inverse image of 2 is b and {b} is not open in the given topology on {a,b} this it is not continuous.

As for the second question. Take an open set in N and look at its inverse image, is it open? That is all you need to do. So what is an open set in the finite complement topology? A set is open if its complement is a finite set. The inverse image of any set is the square roots of the perfect squares in it. eg, the inverse image of the set {2,3,4,5,6,7,8,9} is {1,2,3}.

It may help for you to also prove the following:

let X and Y be topological spaces and f a map from X to Y. Then f is continuous iff the inverse image of any closed set is closed.

Why? Because a closed set in the finite complement topology is precisely a finite set of points, adn the inverse image of a finite set of points is, at most, a FILL IN THE BLANKS
 
  • #26
Oxymoron
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So we are talking about [itex]\mathbb{N}[/itex] with the finite complement topology. This means that a set [itex]U = \mathbb{N}[/itex] is open if its complement [itex]\mathbb{N} \backslash U[/itex] is a finite number of points (or the entire set ie. [itex]U = \oslash[/itex]).

So if I take any subset [itex]U[/itex] of [itex]\mathbb{N}[/itex] then the image of [itex]U[/itex], [itex]f^{-1}(U)[/itex] is empty if [itex]U[/itex] does not contain any perfect squares, (and so [itex]\mathbb{N} \backslash U = \mathbb{N}[/itex] is an infinite number of points - hence closed), or it will be some other subset of [itex]\mathbb{N}[/itex] (since the square root of a positive perfect square is always a positive natural number). And hence the complement [itex]\mathbb{N}\backslash U[/itex] is still an infinite set, and so closed.

Regardless of whether [itex]U[/itex] contains perfect squares or not, the complement of the inverse image [itex]\mathbb{N} \backslash (f^{-1}(U))[/itex] will always contain an infinite number of points. Hence the inverse image of every open subset [itex]U \subseteq \mathbb{N}[/itex] is closed under the finite complement topology.

Therefore the function

[tex]f:\mathbb{N} \rightarrow \mathbb{N}, \, f(x) = x^2[/tex]

is not continuous.


let X and Y be topological spaces and f a map from X to Y. Then f is continuous iff the inverse image of any closed set is closed.

But I thought a function is continuous iff the inverse image of any open set is open? Is this another definition of continuity?
 
  • #27
matt grime
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Oxymoron said:
So we are talking about [itex]\mathbb{N}[/itex] with the finite complement topology. This means that a set [itex]U = \mathbb{N}[/itex] is open if its complement [itex]\mathbb{N} \backslash U[/itex] is a finite number of points (or the entire set ie. [itex]U = \oslash[/itex]).


correct


So if I take any subset [itex]U[/itex] of [itex]\mathbb{N}[/itex] then the image of [itex]U[/itex], [itex]f^{-1}(U)[/itex] is empty if [itex]U[/itex] does not contain any perfect squares,

This is impossible unless U is the empty set. An open set contains all but a finite nubmer of points. In particular it must contain all integers larger than som natural numberr M, eg take M to be the maximum element of the complement of U which exists since the complement of Uis finite.

(and so [itex]\mathbb{N} \backslash U = \mathbb{N}[/itex] is an infinite number of points - hence closed),

I fail to see the deductive reasoning here. What is closed? U or N\U? And do'nt you mean N\f^{-1}(U)?


or it will be some other subset of [itex]\mathbb{N}[/itex] (since the square root of a positive perfect square is always a positive natural number). And hence the complement [itex]\mathbb{N}\backslash U[/itex] is still an infinite set, and so closed.

I'm sorry, again, what is closed f^{-1}(U) or its complement?


Regardless of whether [itex]U[/itex] contains perfect squares or not, the complement of the inverse image [itex]\mathbb{N} \backslash (f^{-1}(U))[/itex] will always contain an infinite number of points. Hence the inverse image of every open subset [itex]U \subseteq \mathbb{N}[/itex] is closed under the finite complement topology.

Therefore the function

[tex]f:\mathbb{N} \rightarrow \mathbb{N}, \, f(x) = x^2[/tex]

is not continuous.


the reasoning is internally inconsisten,and the conclusion wrong, plus you also seem to think that a if S is an infinite set of points that its complement is necessarily closed. This is not true. Consider the even integers in N. This set is infinite but neither it, nor its complement, are open or closed.




But I thought a function is continuous iff the inverse image of any open set is open? Is this another definition of continuity?

No, I told you to prove that a fucntion is continuous iff the inverse image of closed sets are closed, where the definition of continuous is that the inverse images of open sets are open. That is I wanted you to show that

"the inverse images of all open sets are open" is equivalent to the statement "the inverse images of all closed sets are closed"
 
  • #28
Oxymoron
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Thankyou for putting up with me on this Matt - as I struggle to understand this :cry:

Im going to take this one step at a time.

If I use the definition of a continuous function to be

"A function is continuous iff the inverse image of a open set is open"

Then [itex]f:\mathbb{N} \rightarrow \mathbb{N}[/itex] (etc..) is continuous if and only if for every open set [itex]U[/itex] in [itex]\mathbb{N}[/itex] the inverse image of that set [itex]f^{-1}(U)[/itex] is open in [itex]\mathbb{N}[/itex].

THE CHOICE OF MY OPEN SUBSET IN [itex]\mathbb{N}[/itex]

I want to be able to take any open subset in the natural numbers and show that its inverse image is open in the natural numbers:

Let [itex]U[/itex] be an open subset of [itex]\mathbb{N}[/itex]. So [itex]U[/itex] is a set of natural numbers such that [itex]\mathbb{N} \backslash U[/itex] is finite - by definition of an open set under the finite complement topology.

My query here is that [itex]\mathbb{N} \backslash U[/itex] is just the set of natural numbers with a finite number of points removed. So isn't [itex]\mathbb{N} \backslash U[/itex] still an infinite set? Hence [itex]U[/itex] isn't closed.
 
  • #29
matt grime
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Certaily an open set in the finite complement topology is not closed (when the topological space is not a finite set), but, as the saying goes, what has that got to do with the price of fish?

Here is another characterization of sets in the open sets in this topology.

A set S in N with the finite complement topology is open if there is an integer M such that n>M implies n is in S (or S is empty). Ie we can split S into two parts

U and V where V={M+1,M+2,M+3,.....} and U is some subset of {1,2,..,M-1,M}


Now can you see if f is continuous?


ANd you really should try and prove the equivalence of the open v. closed criterion for continuity since it is easy. here are some thoughts:

if f:X to Y is a map of topological spaces and V is an open set, V' its complement is closed.

f^{-1}(Y)=X

if S and T are two subsets of Y then f^{-1}(SuT) is f^{-1}(S)uf^{-1}(T)

further if S and T are disjoint so are f^{-1}(S) and f^{-1}(T)

if S is open and f is continuous then the inverse image of S is open. if T is S complement then what can you say?
 
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  • #30
matt grime
Science Advisor
Homework Helper
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Oxymoron said:
My query here is that [itex]\mathbb{N} \backslash U[/itex] is just the set of natural numbers with a finite number of points removed.


No, U was open. thus N\U is a finite set of natural numbers. It is not the set of natural numbers with a finite number of poitns removed.

It really is easier to think of the closed sets, you know, A set is closd in the finite complement topology iff it is finite or the whole space or the empty set.

The inverse image of the whole space is the whole space and the inverse image of any finite set or the empty set is either empty or the set of square roots of the perfect squares in the original closed set and is thus fniite. Hence the inverse image of a closed set is closed.
 
  • #31
Oxymoron
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It really is easier to think of the closed sets, you know, A set is closd in the finite complement topology iff it is finite or the whole space or the empty set.

Thankyou Matt. Using closed sets the continuity is easier to prove because of the topology.
 

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