Equipotency doubts

Original post by alex1114

Hi! I came across this question and i don´t know how to answer it.

Q. Prove that [0,1] and (0,1) are equipotent by defining a bijection.

And the theory behind this ( Cardinality and Numerability) seems a bit confusing to me.. so if someone could explain this to me i would really appreciate it! Thanks!

I'm not sure how better to explain it than "prove that two sets are equipotent by defining a bijection between them" :P is there anything in particular you don't understand?

By the way, this question is *much* easier if you are allowed to use the Schröder-Bernstein theorem: that two sets are bijective if there are injections both ways. (However, I think the wording of the question precludes that, because it wants you to define a bijection.)

Note also that no such bijection can be continuous, because the continuous image of a compact set is compact, but (0,1) is not compact while [0,1] is.

Reply 2

Original post by alex1114

Hi! I came across this question and i don´t know how to answer it.

Q. Prove that [0,1] and (0,1) are equipotent by defining a bijection.

And the theory behind this ( Cardinality and Numerability) seems a bit confusing to me.. so if someone could explain this to me i would really appreciate it! Thanks!

By the way, I've got a suitable bijection explicitly now. It's related to the fact that

\mathbb{N}

is bijective with

\{2, 3, \dots, \}

.

Reply 3

OP

The trouble is, that i don't know how i can make that bijection.. Completely lost. ( By the way, this is not homework, it is just a random question i found to do, and to be honest , i would be really grateful if you could show me how to tackle it) Many thanks

Reply 4

Original post by alex1114

The trouble is, that i don't know how i can make that bijection.. Completely lost. ( By the way, this is not homework, it is just a random question i found to do, and to be honest , i would be really grateful if you could show me how to tackle it) Many thanks

OK. You need a way to insert two elements into an infinite set. An easier one: can you biject

\{1, 2, \dots \}

with

\{2, 3, \dots \}

?

Reply 5

OP

Original post by Smaug123

OK. You need a way to insert two elements into an infinite set. An easier one: can you biject

\{1, 2, \dots \}

with

\{2, 3, \dots \}

?

I am sorry if i am looking foolish by asking these sort of questions, but, to make a bijection do i have to define it through a function ? e.g a bijection between N( natural numbers) and even is n----> 2n so then 1-->2 ,
2--->4 and so on...

I know that a bijection is when each and every element in the original set is paired with one element of the other set ( one to one functions?), but, how can i establish a bijection between two sets? Or is a bijection of

with

simply {(1,2),(2,3)....} ?
Or to make a bijection between those sets i would have to say that n ---> n+1 so that 1--->2 2---->3 ??

Reply 6

Original post by alex1114

I am sorry if i am looking foolish by asking these sort of questions, but, to make a bijection do i have to define it through a function ? e.g a bijection between N( natural numbers) and even is n----> 2n so then 1-->2 ,
2--->4 and so on...

I know that a bijection is when each and every element in the original set is paired with one element of the other set ( one to one functions?), but, how can i establish a bijection between two sets? Or is a bijection of

with

simply {(1,2),(2,3)....} ?
Or to make a bijection between those sets i would have to say that n ---> n+1 so that 1--->2 2---->3 ??

Yes. A *bijection* is a function which is both injective and surjective. That is, it is a one-to-one mapping from all of its domain to all of its range.

Strictly, the bijection you refer to is indeed the function

m \mapsto m+1

. However, it is perfectly clear what you mean when you write {(1,2),(2,3),…}.

OK. You have a way of "adding elements to a countable-infinite set by a bijection" (which you've just demonstrated by adding the element 1 to the list {2, 3, …}). Does this help you to extend (0,1) to [0,1] by a bijection?

Reply 7

OP

Original post by Smaug123

Yes. A *bijection* is a function which is both injective and surjective. That is, it is a one-to-one mapping from all of its domain to all of its range.

Strictly, the bijection you refer to is indeed the function

m \mapsto m+1

. However, it is perfectly clear what you mean when you write {(1,2),(2,3),…}.

OK. You have a way of "adding elements to a countable-infinite set by a bijection" (which you've just demonstrated by adding the element 1 to the list {2, 3, …}). Does this help you to extend (0,1) to [0,1] by a bijection?

To be honest, i don´t know how to relate to an other set elements which are not included in the set (0 and 1 not included in (0,1) ) so i am confused about that

Reply 8

DFranklin

18

Original post by alex1114

To be honest, i don´t know how to relate to an other set elements which are not included in the set (0 and 1 not included in (0,1) ) so i am confused about that

Not entirely sure what you mean, but note that you have just related two sets N = {1, 2, 3, ...} and M = {2, 3, ...} where 1 is in N but not in M.

Mapping (0,1) to [0,1] is not actually much harder once you see how to do it, but I certainly found it hard to find the right approach.

Here's one method (which I've tried to break into spoiler protected steps):

Spoiler

Reply 9

OP

Spoiler

bijection between (0,1) and [0,1] ,, f(x) = x U {0,1} is that right then?

Reply 10

Original post by alex1114

Spoiler

bijection between (0,1) and [0,1] ,, f(x) = x U {0,1} is that right then?
It's good practice to always state the domain and range of your functions; that would have told you in this instance that your function isn't of the right form. You're trying to find a function from

(0,1)

to

[0,1]

. Suppose you feed the function

f(x) = x \cup \{0,1\}

with the value

x=\frac{1}{2}

. Your function now spits out the value

\frac{1}{2} \cup \{0,1\}

, which doesn't even make sense as a mathematical construct.

Reply 11

OP

Original post by Smaug123

It's good practice to always state the domain and range of your functions; that would have told you in this instance that your function isn't of the right form. You're trying to find a function from

(0,1)

to

[0,1]

. Suppose you feed the function

f(x) = x \cup \{0,1\}

with the value

x=\frac{1}{2}

. Your function now spits out the value

\frac{1}{2} \cup \{0,1\}

, which doesn't even make sense as a mathematical construct.[/QUOTI

How would it be then?

Reply 12