[0,1] same cardinality as (0,1)

[Mr Davis 97]

To show that two sets have the same cardinality you have to show that there is a bijection between the two. Apparently, one bijection from [0,1] to (0,1) is $f(x) = \left\{ \begin{array}{lr} 1/2 & : x = 0\\ \frac{1}{n+2} & : x = \frac{1}{n}\\ x & : \text{any other value} \end{array} \right.$

My question, if I were tasked with finding this bijection, how would I do it? It seems very non-obvious.

 

[fresh_42]

Good question. An approach I can think of is the following: We want to show that $(0,1) \stackrel{1:1}{\longleftrightarrow} (0,1) \cup \{1,\ldots n\}$. Now we know that $\{n+1,\ldots ,\infty\} \stackrel{1:1}{\longleftrightarrow} \{1,\ldots ,\infty\}$ which can be done by shifting the values. To get them into $[0,1]$ isn't difficult, just take their reciprocal. So all what is left, is to put in the $0$ somewhere and define the rest to be constant.

 

[Mr Davis 97]

Good question. An approach I can think of is the following: We want to show that $(0,1) \stackrel{1:1}{\longleftrightarrow} (0,1) \cup \{1,\ldots n\}$. Now we know that $\{n+1,\ldots ,\infty\} \stackrel{1:1}{\longleftrightarrow} \{1,\ldots ,\infty\}$ which can be done by shifting the values. To get them into $[0,1]$ isn't difficult, just take their reciprocal. So all what is left, is to put in the $0$ somewhere and define the rest to be constant.

Why is $(0,1) \stackrel{1:1}{\longleftrightarrow} (0,1) \cup \{1,\ldots n\}$ what we want to show? How is the RHS the same as $[0,1]$?

 

[fresh_42]

Why is $(0,1) \stackrel{1:1}{\longleftrightarrow} (0,1) \cup \{1,\ldots n\}$ what we want to show? How is the RHS the same as $[0,1]$?

I only wanted to say, that the cardinality of an infinite set won't be changed, if we add a finite number of points. This is slightly more general as the case we need, which is $(0,1) \cup \{0,1\}=[0,1]$ but basically the same principle. As we have to add $2$ elements, the shifting operation is $n \mapsto n+2$ to get space for the additional elements and which is the denominator of the sought bijection.

 

[mfb]

How I would approach it: f(x)=x is good for most points, but we have to "hide" x=0 and x=1 somewhere.
There is a very easy bijection between {0,1,2,3,...} and {2,3,...}: g(n)=n+2. As you can see, the first set has two elements the other one has not but we found a bijection. We just have to find a similar set in (0,1). The reciprocal numbers are suitable.
Start with f(0)=1/2, f(1)=1/3. What about f(1/2)? Well, map it f(1/2)=1/4. Continue: f(1/3)=1/5 and f(1/4)=1/6 and so on. Written generally, this directly gives the bijection you have.

It is not the only one, but it is one with a very short notation.

 

[Mr Davis 97]

How I would approach it: f(x)=x is good for most points, but we have to "hide" x=0 and x=1 somewhere.
There is a very easy bijection between {0,1,2,3,...} and {2,3,...}: g(n)=n+2. As you can see, the first set has two elements the other one has not but we found a bijection. We just have to find a similar set in (0,1). The reciprocal numbers are suitable.
Start with f(0)=1/2, f(1)=1/3. What about f(1/2)? Well, map it f(1/2)=1/4. Continue: f(1/3)=1/5 and f(1/4)=1/6 and so on. Written generally, this directly gives the bijection you have.

It is not the only one, but it is one with a very short notation.

I think it's clear to me now. One more questions: would the best way to show that this is a bijection be to find the inverse, or to show that it is surjective and injective?

 

[fresh_42]

There is no "best" way. Both are reasonably short, but don't forget, that the inverse method needs both sides: $f\circ f^{-1}= \operatorname{id}$ and $f^{-1} \circ f = \operatorname{id}$. Surjectivity is short, injectivity requires an argument, but isn't complicated either.

 

[SSequence]

I think perhaps one way might be to see how to create a bijection between (0,1] and (0,1). So, for example, we could have:
f(1) = 1/2
f(1/2) = 1/3
f(1/3) = 1/4
f(1/4) = 1/5
... and so on.
Apart from the values of the form 1/n where n ≥ 1, we would have f(x)=x.

I suppose post#5 is very similar. But I wanted to emphasize a bit (like mentioned in post#4) that adding any finite number of elements wouldn't change the cardinality of (0,1) because we could create more backward chains like the one mentioned above. But we would have to be careful that the backward chains remain completely separate from each other.

In fact, of course adding a countable infinity of elements wouldn't change the cardinality of (0,1) either (create a countable infinity of separate backward chains). For example adding a set A to (0,1) ... where A={1,2,3,4,5,...}.

==========

I was thinking, how about bijection between (0,1) and (0,∞). If we assuming a well-ordering of [0,1] (and hence also (0,1) ) is possible, then it is fairly easy to see a bijection.

If not, probably something like this should work I think:
create correspondence between (0,1/2] and (0,1]
creater correspondence between (1/2,3/4] and (1,2]
and so on shrinking the intervals further and further.

Though I am guessing this kind of construction would be routine for those familiar with solving problems of this kind.

 

[mfb]

I suppose post#5 is very similar. But I wanted to emphasize a bit (like mentioned in post#4) that adding any finite number of elements wouldn't change the cardinality of (0,1) because we could create more backward chains like the one mentioned above. But we would have to be careful that the backward chains remain completely separate from each other.

You can use the same chain for every countable number of elements. OP had an example for 2 elements.
This is related to Hilbert's hotel.

 

[FactChecker]

That's a clever bijection, but I think there are simpler, more routine, ways. For instance, you could use the bijection f(x) = x/2+1/4 of [0,1] to [1/4, 3/4] ⊂ (0,1) to prove that the cardinality of [0,1] ≤ cardinality of (0,1) and clearly (0,1) ⊂ [0,1] so cardinality of (0,1) ≤ cardinality of [0,1]. Concluding that the cardinalities are equal. It's not as clever, but I think it gets the job done.

 

[SSequence]

You can use the same chain for every countable number of elements. OP had an example for 2 elements.
This is related to Hilbert's hotel.

I guess there would be pretty clever ways to write this kind of bijection so I didn't discount any other method in my wording (but I tried to keep it brief).

But specifically looking at OP's construction, I suppose it is just a little difference in meaning of chain? I plugged in some values from the function in OP. Here is how it goes:
f(0) = 1/2
f(1/2) = 1/4
f(1/4) = 1/6
f(1/6) = 1/8
and so on...

The way I was considering it, I was counting that as a chain. And then for second element plugging the values:
f(1) = 1/3
f(1/3) = 1/5
f(1/5) = 1/7
f(1/7) = 1/9
and so on...

The way I was using the word "chain", I was considering these as two separate chains. I suppose you were referring to chain roughly in the sense of set of values that had to be altered (or something similar).

Really doesn't matter much though I guess.