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March 8th, 2017, 10:28 AM   #11
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Quote:
Originally Posted by zylo View Post
You missed the point. Every convergent sequence in R is mapped to a convergent sequence in (0,1) except the one which maps to the convergent sequence whose limit is 0, which is not in (0,1), but a border point which doesn't map to R.

The conclusion is not a puzzle. It is a trick question.
Please turn to Rudin, Principles of Mathematical Analysis, Second Edition (I believe this is the version you're working from), Chapter 3, "Numerical Sequences and Series," page 41. We find:

Quote:
Originally Posted by Rudin
A sequence $\{p_n\}$ in a metric space $X$ is said to converge if there is a point $p \in X$ with the following property: For every $\epsilon > 0$ there is an integer $N$ such that $n \geq N$ implies that $d(p_n, p) < \epsilon$.
Rudin goes on to say, a couple of sentences later:

Quote:
Originally Posted by Rudin
It might be well to point out that our definition of "convergent sequence" depends not only on $\{p_n\}$ but on $X$; for instance, the sequence $\{\frac{1}{n}\}$ converges in $\mathbb R^1$ (to $0$) but fails to converge is the set of all positive real numbers ...
As you see, Rudin explicitly discussed the case at hand and pointed out that $\{\frac{1}{n}\}$ does not converge in the positive real numbers, or (in our situation) in the metric space $(0,1)$.

Also note that in Chapter 4, Rudin defines continuity for functions in a metric space, and shows (theorem 4.6) that a continuous function preserves limits.

Last edited by Maschke; March 8th, 2017 at 10:40 AM.
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March 8th, 2017, 10:32 AM   #12
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A) You've missed the point described above.
B) No convergent sequence in $\mathbb R$ maps to one that "converges" to zero on $(0,1)$ because such sequences in $\mathbb R$ diverge to $-\infty$.
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March 11th, 2017, 12:07 AM   #13
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Quote:
Originally Posted by zylo View Post
Can you map (0,1) to R bijectively and continuously both ways?
How?

Reference: Maschke, Post #1
Relatively Open and Closed Sets
No:

1/n converges to a limit (which is not in (0,1)).

f(1/n) converges to a limit (which is in R because R is complete). ->

1/n converges to a limit in (0,1). Contradiction.
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March 11th, 2017, 05:29 AM   #14
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Quote:
Originally Posted by zylo View Post
1/n converges to a limit (which is not in (0,1)).

f(1/n) converges to a limit (which is in R because R is complete).
This is what is so infuriating about you. Both of these statements have been shown to be false in this thread. For the first, Maschke highlighted chapter and verse of your text. For the second, I pointed out that as $\displaystyle n \to \infty$ we find that $\displaystyle f\left(\frac1n\right) \to -\infty$. And both sequences are thus divergent.

Why would you continue to peddle the same fiction?

Last edited by skipjack; March 11th, 2017 at 10:30 AM.
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March 18th, 2017, 07:49 AM   #15
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Quote:
Originally Posted by zylo View Post
Can you map (0,1) to R bijectively and continuously (homeomorphically) both ways?
No

Quote:
Originally Posted by zylo View Post

Assume f homeomorphic from (a,b) to R, a<b.


For any $\displaystyle \delta$, y=f(x) is bounded on (a+$\displaystyle \delta$, b-$\displaystyle \delta$) $\displaystyle \rightarrow$

$\displaystyle \lim_{y\rightarrow -\infty}f^{-1}(y) = a$ and $\displaystyle \lim_{y\rightarrow \infty}f^{-1}(y) = b$ because f$\displaystyle ^{-1}$ continuous on R. $\displaystyle \rightarrow$

f is not homeomorphic.


Example: y = tanx is not homeomorphic from (-$\displaystyle \pi $/2,$\displaystyle \pi $/2) to R.
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