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$S$ denotes the set of real numbers strictly between $0$ and $1$. That is, $S = \{x \in R\mid 0 < x < 1\}$.

Let $U = \{x \in R\mid 0 < x < 2\}$. Prove that $S$ and $U$ have the same cardinality.

I am completely lost. How do I approach this problem?

Michael Hardy
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Snowyace
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3 Answers3

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Recall that two sets $X$ and $Y$ are of the same cardinality if there exists a bijection between them. That is, we may construct a 1-1 and onto function from $X$ to $Y$. There is certainly such a map between your two sets (in fact there a uncountably many) but one obvious one. I'll give a hint: its a very simple map and it takes the ends of the interval in $S$ to the ends of the interval in $U$.

Joseph Zambrano
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  • Thank you for the hint.

    I'm guessing the function is f(x) = 2x because the interval of S could be mapped to the interval of U.

    I'm interpreting S as the domain and U as the co-domain.

     – Snowyace Mar 10 '14 at 01:15
  • You got it. Now a bijection will have a well defined inverse as well. I'm sure you realize that you can work the other direction as well and in this case you will get $f(x)=1/2x$. Now try to show that it is in fact a bijection (by showing 1-1 and onto). – Joseph Zambrano Mar 10 '14 at 01:32

  • I think I'm starting to understand it.

    For another problem: $$ Let V = {x ∈ R| 2 < x < 5}. $$

    Prove that S and V have the same cardinality.

    I came up with the function f(x) = (3/2)x + 2 through connecting the points of domain. f(0) = 2 and f(2) = 5 then used point slope form. 

Then I showed correspondence of f: S-> V through normal methods of showing its 1-to-1 and onto.
    – Snowyace Mar 10 '14 at 01:51
  • @Joseph $1/2x$ does not mean $(1/2)x$; it means $1/(2x)$. You should probably write $x/2$ or $\frac x2$ ($\frac x2$). – MJD Mar 10 '14 at 02:01
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Hint: The function is : f(x) = 2x

DeepSea
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Two sets have the same cardinality if there exists a bijection between them; that is, a function that is a one-to-one correspondence between the two sets.

All you have to do is prove that (1)$f(a)=f(b) \implies a=b \forall a \in S$ and (2) $\forall u \in U, \exists s \in S$ such that $f(s)=u$. This should be fairly straightforward, and for (2), I suggest finding an expression $s$ in terms of $u$ for any arbitrary $u \in U$. [$s=u/2$.]

William Chang
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