Rectangle question and closure of the interior?

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SUMMARY

The discussion focuses on proving that for a rectangle Q = [a1,b1] x ... x [an,bn], the set Q is equal to the closure of its interior, denoted as Cl(Int(Q)). The user seeks to demonstrate that Cl(Int(Q)) = Int(Int(Q)) U bd(Int(Q)). Key steps include establishing that Int(A x B) = Int(A) x Int(B) and identifying the boundary points of Int(Q) as {a1, b1, ..., an, bn}. The rigorous proof involves explicitly writing out the definitions of interior and boundary in set theory notation.

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  • Understanding of set theory notation
  • Familiarity with the concepts of interior and boundary in topology
  • Knowledge of Cartesian products of sets
  • Basic principles of closure in topological spaces
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Buri
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The question says:

Show that if Q = [a1,b1]x...x[an,bn] is a rectangle, the Q equals the closure of Int Q.

The definition of closure that I have is Cl(A) = int(A) U bd(A). So I'd like to show that Cl(int(Q)) = int(int(Q)) U bd(int(Q)).

But this just seems to be obvious to me which just makes it hard to prove - I just don't know what to write. Any hints/ideas on how to prove this rigorously?

EDIT!

I guess I'd have to show something like:

Int(AxB) = Int(A)xInt(B)

And then I guess, I'd make a claim that bd(int(Q)) = {a1,b1,...,an,bn} and prove this by showing that no other boundary points exist?

Questions like this I always find hard.
 
Last edited:
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You are being asked to show that Q = Cl (IntQ)
You know how to define Int and Bd, so write out the definitions explicitly (in set theory notation) to show that they coincide.
 

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