Is 1/x a homeomrphism from x>0 to y>0? Is 1/x a homeomrphism from x>0 to y>0? No. $\displaystyle y = \lim_{x\rightarrow \infty}1/x = 0.$ R is complete. Definition a) A function f: (M1) →(M2) is homeomorphic (bicontinuous) if it has the following. properties: f is a bijection (onetoone and onto), f is continuous, the inverse function f$\displaystyle ^{−1}$ is continuous. This definition implies M1 and M2 are metric spaces (distance defined). 
Good God Almighty why are you starting yet another thread pushing your same incorrect idea? How on earth can x > 0 and y > 0 NOT be homeomorphic? They're the same exact set described using a different variable. Would you falsely claim that the identity function on the positive reals is not a homeomorphism? After all if f(x) = x is the identity function, then $\displaystyle \lim_{x \to 0} x = 0$, which is then subject to your same objection. So I ask you: Is the identity function a homeomorphism from the positive reals to the positive reals? This all comes down to the "puzzler" I tossed out a while back. Continuous functions do not necessarily preserve completeness. When you grok this you will be enlightened. 
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It occurs to me that there's a better way to explain this. What the tshirt should really say is: Continuous functions don't necessarily preserve Cauchy sequences. So the sequence $(\frac{1}{n})$ is Cauchy, and does not attain a limit in $(0,1)$, showing that $(0,1)$ is not complete. If we apply the continuous function $f(x) = \frac{1}{x}$ to that sequence, we get the sequence $1, 2, 3, 4, ...$ which does not converge in the reals. Yet the reals are complete, so what happened? Well, $1, 2, 3, 4, \dots$ is not Cauchy and therefore its nonconvergence doesn't count against the completeness of the reals. 
I suspect that he refuses to accept that a Cauchy sequence doesn't necessarily converge in a given domain. In other words, he claims that $\{\frac1n\}$ converges in $(0,\infty)$. This then causes problems down the line. As usual, if you don't accept the standard definitions, you can't use the theorems that use those definitions. 
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$\displaystyle \lim_{x\rightarrow a}1/x$ $\displaystyle \epsilon$ y>0 for every a in x>0, in which case answer to OP is yes. From that point of view, tan is homeomorphic on ($\displaystyle \pi$/2,$\displaystyle \pi$/2). Don't recall anyone making that, or any other, rational argument, for or against OP. So 1/x did serve a purpose, to distill a question to its more transparent essence. Many thanks to all participants, whose responses I found quite motivational. So what is answer to OP? Yes or No. It reminds me of a real number being defined by a decimal for all n, or lim as n $\displaystyle \rightarrow \infty$. 
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Yes of my previous post also holds. 
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The positive reals are a set. You know how set theory works. Everything we know about a set depends on its members. There are no "secret numbers" clinging to the outside of a set like barnacles. 
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