Counter-Example to Showing T' Is Finer than T

[micronemesis]

let X and X' be two single sets in two topologies T and T' respectively.A standard textbook asks to show that if identity function i:X' →X is continuous then T' is finer than T.Is the example below a valid counter example to this argument?

Homework Equations

3. The example goes like this .Consider γ={1,2,3,4,5} and T={∅,γ,{1,2},{1},{3},{1,2,3},{1,3}} and T'={∅,γ,{1},{1,4}}.X={1,2} and X'={1}.so the subspace topology on X due to T i.e Tx={∅,X,{1}} and T'x'={∅,X'}.It is obvious that inverse of each open element in X' is open in X, but T' is not finer than T.

 

[Samy_A]

let X and X' be two single sets in two topologies T and T' respectively.A standard textbook asks to show that if identity function i:X' →X is continuous then T' is finer than T.Is the example below a valid counter example to this argument?

Homework Equations

3. The example goes like this .Consider γ={1,2,3,4,5} and T={∅,γ,{1,2},{1},{3},{1,2,3},{1,3}} and T'={∅,γ,{1},{1,4}}.X={1,2} and X'={1}.so the subspace topology on X due to T i.e Tx={∅,X,{1}} and T'x'={∅,X'}.It is obvious that inverse of each open element in X' is open in X, but T' is not finer than T.

Is i:X' →X continuous?

 

[WWGD]

But usually the identity function is defined from a space to itself, or at least $X' \subset X$.

 

[micronemesis]

But usually the identity function is defined from a space to itself, or at least $X' \subset X$.

that is clear as day light.i think that this example even if it satisfies the premise of the question it does not satisfy its implication.but the question being from a very very well acclaimed book,i wonder whether i got it wrong and if i got it wrong then there has to be some serious issues the way which i am doing this subject.

 

[Samy_A]

that is clear as day light.i think that this example even if it satisfies the premise of the question it does not satisfy its implication.but the question being from a very very well acclaimed book,i wonder whether i got it wrong and if i got it wrong then there has to be some serious issues the way which i am doing this subject.

May I repeat the question: is i:X' →X continuous?
In post #1 you write "It is obvious that inverse of each open element in X' is open in X". But what you have to look at are the inverses of open elements in X.

 

[micronemesis]

May I repeat the question: is i:X' →X continuous?
In post #1 you write "It is obvious that inverse of each open element in X' is open in X". But what you have to look at are the inverses of open elements in X.

i am really happy that you replied ;didn't expect anyone to be frank; and from your reply it is obvious that you have understood the question.i actually misplaced that particular phrase.my intention was to put it as "inverse of every open set in X is open in X' ", as you can see i [-1]{1}={1}, i [-1]{1,2}={1}, i [-1]Φ=Φ all are open in X'.i noticed this mistake early but being entirely new to this kind of platform i didn't know how to correct it.appreciate your effort.if possible please follow this question to its conclusion.also i am not able to put the inverse function of i in its proper representaion. please read i [-1] as " i inverse" whenever encountered.

 

[Samy_A]

i am really happy that you replied and from your reply it is obvious that you have understood the question.i actually misplaced that particular phrase.my intention was to put it as "inverse of every open set in X is open in X' ", as you can see [-1]{1}={1},[-1]{1,2}={1},[-1]Φ=Φ all are open in X'.appreciate your effort.if possible please follow this question to its conclusion.

Ok, you fixed the OP, it was a typo.

But you are dealing here with two different sets, $X' \subset X$. How do you define "finer" in that case? Obviously $\{1,2\}$ can't be in T', as 2 isn't an element of $X'$.
Could you post the precise question from the book?

What is generally claimed is that if you have a set Y with two topologies, S and S', then the identity map
$I: Y,S \to Y,S'$ is continuous if and only if $S' \subseteq S$.

I you want to extend this to identity maps between real subsets $X' \subset X$, then you will have to compare the topology on X' with the topology induced on X' by the topology on X.

 

[micronemesis]

the question goes like this exactly as follows." let X and X' denote two single set in topologies T and T', respectively.let i:X' →X be the identity function.
p: i is continuous.
q: T' is finer than T. "
first of all do you agree on following points ?
(1) T and T' are two topologies on same set ; in my case on {1,2,3,4,5} is obvious from the question.so discussing the notion of "fineness" between T and T' is valid.

your suggestion " But you are dealing here with two different sets, X&X' X′⊂X. How do you define "finer" in that case? " seems to suggest that you are considering the notion of " fineness" between Tx and T'x' rather than doing the same between T and T' as the question literally suggests.and for your reference the question can be found in the following link "http://dbfin.com/topology/munkres/chapter-2/section-18-continuous-functions/problem-3-solution/"

 

[Samy_A]

the question goes like this exactly as follows." let X and X' denote two single set in topologies T and T', respectively.let i:X' →X be the identity function.
p: i is continuous.
q: T' is finer than T. "
first of all do you agree on following points ?
(1) T and T' are two topologies on same set ; in my case on {1,2,3,4,5} is obvious from the question.so discussing the notion of "fineness" between T and T' is valid.

your suggestion " But you are dealing here with two different sets, X&X' X′⊂X. How do you define "finer" in that case? " seems to suggest that you are considering the notion of " fineness" between Tx and T'x' rather than doing the same between T and T' as the question literally suggests.and for your reference the question can be found in the following link "http://dbfin.com/topology/munkres/chapter-2/section-18-continuous-functions/problem-3-solution/"

The question in your link is: "Let X and X' denote a single set in the two topologies ..."
Not two sets as you write.

As sets, X=X', so your counterexample is moot.

 

[micronemesis]

The question in your link is: "Let X and X' denote a single set in the two topologies ..."
Not two sets as you write.

In my case X={1,2}∈T and X'={1}∈T'.So the sets which i took are open in the topologies that i considered

 

[Samy_A]

In my case X={1,2}∈T and X'={1}∈T'.So the sets which i took are open in the topologies that i considered

Yes, you have two sets. The books talks about one set (with two topologies).

For reference, this is the full question in your link:

So no, you don't have a valid counterexample to the correct claim in this exercise.

 

[micronemesis]

Yes, you have two sets. The books talks about one set (with two topologies).

For reference, this is the full question in your link:
View attachment 95911

So no, you don't have a valid counterexample to the correct claim in the book.

the one set that the book talks about in my case is {1,2,3,4,5} , T and T' being two different topologies on it and {1,2} and {1} are two single sets in T and T' respectively.

 

[Samy_A]

the one set that the book talks about in my case is {1,2,3,4,5} , T and T' being two different topologies on it and {1,2} and {1} are two single sets in T and T' respectively.

The book asks you to consider i: X' → X. In your case, as sets, X=X'={1,2,3,4,5}
Now, you can check whether the function i is continuous or not with the topologies T' on X' and T on X.
Nowhere does the exercise refer to an "identity function" between two different sets.

 

[micronemesis]

The book aks you to consider i: X' → X. In your case, as sets, X=X'={1,2,3,4,5}
Now, you can check whether the function i is continuous or not with the topologies T' on X' and T on X.
Nowhere does the exercise refer two an "identity function" between two different sets.

so you mean to say whatever be the context, if one phrases "f:A→B be the identity function " then it naturally means that A=B.

 

[Samy_A]

so you mean to say whatever be the context, if one phrases "f:A→B be the identity function " then it naturally means that A=B.

f could be an inclusion, when A is a subset of B. I guess one could call that the identity function.
But in this case, in this exercise, it is clearly stated that X and X' denote a single set (with two topologies).

 

[micronemesis]

f could be an inclusion, when A is a subset of B. I guess one could call that the identity function.
But in this case, in this exercise, it is clearly stated that X and X' denote a single set (with two topologies).

finally got it.i thought what the book meant was that X and X' are just two single set members of T and T'.But now I feel like your explanation is correct as the book is saying to consider a particular set as two different set in the context of two different topologies.i feel satisfied with your explanation.ironically the problem that got me into this mess is my micronemesis i.e english.anyway appreciate your genuineness and benignness