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Show that $([0,1]]\cup(2,3],<)$ has the same order type as $([0,1],<)$.

Workings:

I'm not entirely sure what to do with this question. I know I have $f(a)→f(b)$, and I need a function $f(x)$ for this but I can't think of anything.

Any help will be aprreciated

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Define $g(x) = x$ for $x \in [0,1]$ and $g(x) = x-1$ for $x \in (2,3]$.

This defines an order isomorphsim between $[0,1] \cup (2,3]$ and $[0,2]$. The latter clearly has the same order type as $[0,1]$ using $h(x) = \frac{x}{2}$.

Now use that having the same order type is an equivalence relation, or compose $h \circ g$, to get a direct order isomorphism:

$\phi(x) = \frac{x}{2}$ for $x \in [0,1]$ and $\phi(x) = \frac{x-1}{2}$ for $x \in (2,3]$.

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  • $\begingroup$ Thanks a lot. That makes sense/ $\endgroup$ – CoolNewFriends Jan 29 '17 at 20:34

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