# Gleeful bane

1. Feb 14, 2016

### 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?

2. Relevant 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.

Last edited by a moderator: Feb 14, 2016
2. Feb 14, 2016

### Samy_A

Is i:X' →X continuous?

3. Feb 14, 2016

### WWGD

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

4. Feb 14, 2016

### micronemesis

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.

5. Feb 14, 2016

### Samy_A

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.

Last edited: Feb 14, 2016
6. Feb 15, 2016

### micronemesis

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.

7. Feb 15, 2016

### Samy_A

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.

Last edited: Feb 15, 2016
8. Feb 15, 2016

### 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/"

9. Feb 15, 2016

### Samy_A

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.

10. Feb 15, 2016

### micronemesis

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

11. Feb 15, 2016

### Samy_A

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

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

12. Feb 15, 2016

### micronemesis

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.

13. Feb 15, 2016

### Samy_A

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.

Last edited: Feb 15, 2016
14. Feb 15, 2016

### micronemesis

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.

15. Feb 15, 2016

### Samy_A

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).

16. Feb 15, 2016

### micronemesis

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