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I know the line segment have a infinite number of points, but i know that exist different kinds of infinity ( $\aleph_0 $). My question is there same number of points on segment of line and entire line. If you know some good book or article about this topic i will be great full.

Sorry for bad English.

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up vote 2 down vote accepted

If you mean the cardinality of $L$, when you say "number" of points on $L$, then the number of points on a segment of a line and the entire line is same.

For example, consider $f:(-1,1) \to \mathbb{R}$ defined as $f(x) = \tan {\frac{\pi x}{2}}$. $f(x)$ is both one-one and onto.

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Are you sure there is a continuous function from a closed interval which is a bijection with $\mathbb R$? I am fairly certain that there is none. – Asaf Karagila Dec 15 '12 at 19:11
Thanks Andre Nicolas. I have fixed it now. @AsafKaragila: Hmm... I am thinking about it :) – Isomorphism Dec 15 '12 at 19:17
@AsafKaragila: I dont think there exists a continuous function from a bounded closed interval to $\mathbb{R}$, simply because $R$ is open and $f^{-1}(\mathbb{R})$ will be forced to be open. But a closed and bounded interval cannot be open! – Isomorphism Dec 15 '12 at 19:24
That's the wrong argument. $[a,b]$ is open relative to itself. But the interval is compact so a continuous image would be compact and in $\mathbb R$ that means bounded. One could extend $\mathbb R$ by adding $\pm\infty$ points, but that is no longer a real valued function, is it now? – Asaf Karagila Dec 15 '12 at 19:31
But so what? Nobody seems to claim or require that the function is continuous anyway. – Henning Makholm Dec 15 '12 at 19:37

Assuming that the entire line you mean $\mathbb R$ (the real numbers) and the line segment you mean some interval, e.g. $[0,1]$, then there are the same number of points on the two sets.

That is to say there is a bijection between $\mathbb R$ and $[0,1]$. This is discussed in Bijection from finite (closed) segment of real line to whole real line.

I should add that the cardinality (size of infinity) of $\mathbb R$ is strictly greater than $\aleph_0$, and in fact can be calculated to be $2^{\aleph_0}$. Namely, given that $\mathbb N$ has size $\aleph_0$ then the real numbers have the same size as the set $\{A\mid A\subseteq\mathbb N\}$, which can be calculated to be $2^{\aleph_0}$.

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Is any Euclidean space have cardinality of $ 2^{\aleph_0} $ ? – Ubavic Dec 15 '12 at 19:28
Nikola, if by Euclidean space you mean $\mathbb R^n$ then yes. In fact even larger spaces can have this cardinality. For example the space of all polynomials, and even all power series over $\mathbb R$ still have the same cardinality. – Asaf Karagila Dec 15 '12 at 19:32
Thanks for yours responses. – Ubavic Dec 15 '12 at 19:46

If by 'segment of line' you mean 'an interval within an interval', then yes, both have the same cardinality (number of points), the cardinality of the real numbers. Reason: you can easily map the smaller interval (convex subset of a larger convex subset) bijectively to the larger interval (assuming for simplicity the intervals are closed (contain their end points), map end points to end points and 'stretch' as you would if you stretched a smaller rubber band to match a larger rubber band lying under it). As to sources to read, you can start by checking 'cardinality of the continuum' on wiki; or google Munkres, Topology, of which you'll find copies online, and read the first chapter. Any decent analysis book though will cover this.

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