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The intervals $[0,1]$ and $[3,5]$ are equivalent.

My proof goes like this.

Proof. To show that the two sets are equivalent, we should show a bijection between them. Consider the function $f:[0,1] \to [3,5]$ such that $f(x)=2x+3$.

Since $f$ is linear, it is bijective. Therefore, the intervals $[0,1]$ and $[3,5]$ are equivalent.

I am not sure if this is right. Assuming that this is correct, is it better to show that $f$ is reflexive, symmetric, and transitive?

If this is wrong, what is the right way of proving?

Sorry if this is so simple.

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Hindi ko po alam
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  • In set theory, two sets are equivalent if and only if they have the same elements. Therefore, $[3,5]$ and $[0,1]$ are not equal. The correct term to use is isomorphic; bijections are isomorphisms in the category of sets.

    As far as your proof goes, your function is fine, but it's not justified correctly. Why is it bijective? Try appealing to the intermediate value theorem.

     – MathematicsStudent1122 Nov 27 '16 at 05:49
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    As far as I know, set A is equivalent to set B if there exists a bijection between A and B. I'm not talking about equal sets. – Hindi ko po alam Nov 27 '16 at 06:07
  • @MathematicsStudent1122: You are confusing "equal" and "equivalent"; the OP's use of terms is correct. For instance, $\Bbb N$ and $\Bbb Z$ are not equal, but are equivalent. – Alex M. Nov 27 '16 at 10:16
  • You may use the words "equipotent" or "equinumerous" for the relationship you deal with, equivalent is very non-standard. Your function is fine, and if you have a result stating that an affine (probably what you call linear, but that function isn't actual linear) function is a bijection between it's domain and range, I would be happy if you just added a argument about using (and why you can use) it. And it is neither reflexive nor transitive (I don't even think those terms can be defined meaningfully when the domain and co-domain are different as here) so that is just meaningless. – Henrik supports the community Nov 27 '16 at 10:18

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The function is not symmetric since there is the point (0,3) but not the point (3,0) because f is not even defined in 3. Your argument is correct, if you're talking about equivalence as topological spaces (showing there's a continuous bijection between both spaces). May be you could argument that it is bijective because it has a left and right inverse: g(x)=(x-3)/2 and continuity is clear because it is a composition of continuous functions.

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