Show that the set $ [-1, \infty )$ has cardinality $ c \ \ $

Asked Jul 17 '17 at 15:21

Active Jul 17 '17 at 15:39

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Show that the set $ [-1, \infty )$ has cardinality of the continuum. Draw the bijection.

Answer: If I consider the map $ f: [-1, \infty ) \rightarrow (-\infty, \infty) $ by $ f(x)=\log(1+x) $.

This is one-one and onto.

Hence the proof.

But I have doubts using $ f(x)= \log(1+x) $ because it is not defined at $ x=-1 $.

Any help is there ?

edited Jul 17 '17 at 15:39

asked Jul 17 '17 at 15:21

MAS

4

Your $f$ is a bijection $(-1,\infty) \to (-\infty,\infty)$. In order to get a bijection $[-1,\infty) \to (-1,\infty)$, you can send $1/n \mapsto 1/(n+1)$ for integers $n \ge 1$, and send $-1 \mapsto 1$, while leaving everything else fixed. There might be a simpler way to construct a bijection directly from $[-1,\infty)$ to $(-\infty,\infty)$ instead of composing two bijections. – angryavian Jul 17 '17 at 15:28
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Your $f$ is undefined at $x=-1$. – Thomas Andrews Jul 17 '17 at 15:28
You can't get a continuous bijection between the two. – Thomas Andrews Jul 17 '17 at 15:29
how can i draw the bijection – MAS Jul 17 '17 at 15:30
@angryavian What you wrote is an answer. Why don't you post it as such? – José Carlos Santos Jul 17 '17 at 15:31
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Your propositsion works with $(-1,\infty)$. @angryavian gave you a desired bijection. – AlvinL Jul 17 '17 at 15:31
i said in the question that I have to draw bijection – MAS Jul 17 '17 at 15:32
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We'd like to know what that means. Defining an explicit mapping is much more desirable than "drawing". Perhaps, they are synonyms. – AlvinL Jul 17 '17 at 15:33

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