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I am sitting with a task where I have to prove the following:

Claim: Every subset of $\mathbb{R}$, that contains an interval $I$ with $a < b$, has the same cardinality as $\mathbb{R}$.

So I think that I should prove that there exist a bijection from $I$ to $\mathbb{R}$? I'm kinda lost, and don't know how to start.

There is a lemma 3 in the book saying: Let $a,b \in \mathbb{R}$ with $-\infty\neq a < b \neq \infty$. There exist a $f\colon ]-1; 1[ \to ]a; b[$ that is bijektiv. The intervals $]-1; 1[$ and $]a; b[$ has same cardinality.

Another lemma 4 says: $f\colon ]0; 1[ \to ]1; \infty[$, $x\to \frac{1}{x}$, is bijective. The intervals $f\colon ]0; 1[ \to ]1; \infty[$ have same cardinality.

(There is an image added to lemma4)

enter image description here

The above interval $]-1; 1[$ has same cardinality as $\mathbb{R}$, and the $f$ is bijective Any help is highly appreciated

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Let $A$ be a subset that contains a non-trivial interval $I$. Then $\Bbb R=_c I\leq _c A\leq _c\Bbb R$. – Git Gud Oct 16 '13 at 11:44
Are you allowed to use the fact that cardinalities are ordered? That is, do you know that if a subset of $\mathbb{R}$ has a cardinality "not less than" $|\mathbb{R}|$ then it has cardinality equal to $|\mathbb{R}|$? – Eric Stucky Oct 16 '13 at 11:59
Hi. Yes I think I am. Isn't that what the Bernsteins-Schröders sentence says? – user1960836 Oct 16 '13 at 12:02


First prove that any two non-empty open intervals have the same cardinality.

Second, pass now to use the nice interval $\;(-\pi/2\,,\,\pi/2)\;$ and a rather nice, simple trigonometric function to show equipotency with $\;\Bbb R\;$

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There is also a simple rational function that maps $(-1,1)$ to $\mathbb R$ bijectively. – lhf Oct 16 '13 at 11:51
Thanks for the replies people. I am really not good at proofs in maths. That's why I have paid for the course that I am following, so that I can learn more. The posts you have provided so far, doesn't help me more than the lemmas provided, which I can't use to my advantage. Please don't understand this in a negative way, I just don't know how else to say it. So can anyone be a little more specific, or maybe with examples? Thanks – user1960836 Oct 16 '13 at 11:56

For a bijection from $I$ to $\mathbb R$ chose $$f: (a,b) \to (-1,1), \qquad x\mapsto 2\frac{x-a}{b-a} - 1$$ which is (obviously) bijective. Then $$g: (-1,1) \to \mathbb R, \qquad \begin{cases}\frac{1}{x} - 1&x\in(0,1)\\0&x=0\\\frac{1}{x} + 1&x\in(-1,0)\end{cases}$$ (Taken from Lemma 4) $$h := g\circ f : (a,b) \to\mathbb R$$ is bijective as a composition of bijective maps.

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At this part: f:(a,b)→(−1,1),x↦x−ab−a−1. How did you get to -1 at: x↦x−ab−a−1? Also... What is the conclusion on this: h:=g∘f:(a,b)→R? Why do you do that part? – user1960836 Oct 16 '13 at 13:01
Please use MathJax, your text is very hard to read ($\$\text{\frac\{a\}\{b\}}\$$ = $\frac a b$) Second, $f(a) = -1, f(b) = 1$ - since $f$ is linear we know by mean-value theorem $f((a,b)) = (-1,1)$. The $h$ is one bijection from $(a,b)$ onto $\mathbb R$ – AlexR Oct 16 '13 at 13:24
I see that you edited the f function and added the number 2 suddenly? And thank you very much for the help and comments. I am thinking. There should be a comment on the interval (a, b), according to Bernsteins-Schröders sentence? won't it clarify the argument why f is bijective? I know that you can just test it with values, but still, for a more formal argument and proof? – user1960836 Oct 16 '13 at 14:06
I am trying to figure out how you got to that formular of the function f. If it is not too much to ask, I'd be really grateful if you could tell me – user1960836 Oct 16 '13 at 14:23
It's a solution of a simple linear equation: Find $f(x) = ax + b$ s.t. $f(x_0) = y_0, f(x_1) = y_1$. Using the constraints $$f(a) = -1, f(b) = 1$$ you get the desired result. The inverse of $f$ can be found by the same method and $$f^{-1}(-1) = a, f^{-1}(1) = b$$ – AlexR Oct 16 '13 at 15:29

No one has pointed out yet that to prove the statement of the OP (where a subset contains an interval), you're also going to need to invoke the theorem that bears the names Cantor, Schroeder, and Bernstein (in some order or other).

The idea is to construct an injection from $\mathbb{R}$ into the subset $S$ by injecting it into an interval $(a, b)$ inside $S$ (the other answers describe how to do this), and of course we have an injection from $S$ into $\mathbb{R}$ given by subset inclusion. Then apply the Cantor-Schroeder-Bernstein theorem, which says that if there is an injective function $f: A \to B$ and an injective function $g: B \to A$, then there exists a bijection $\phi: A \to B$.

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