Limit Points of Unbounded Interval

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SUMMARY

The closure of the set \( A = [-\infty, 0) \) is definitively \( [-\infty, 0] \). The proof involves demonstrating that all points in \( [-\infty, 0) \) are limit points of \( A \) and that the supremum of \( A \) is \( 0 \), which is included in the closure. To establish equality, one must show that the closure is a subset of \( [-\infty, 0] \) and vice versa, confirming that both sets contain the same elements.

PREREQUISITES
  • Understanding of limit points in topology
  • Familiarity with the concept of closure in metric spaces
  • Knowledge of supremum and infimum in real analysis
  • Basic set theory, particularly subset relations
NEXT STEPS
  • Study the properties of closure in metric spaces
  • Learn about limit points and their significance in topology
  • Explore supremum and infimum in the context of real numbers
  • Investigate set equality and subset proofs in mathematical analysis
USEFUL FOR

Mathematics students, particularly those studying real analysis and topology, as well as educators looking to deepen their understanding of set theory and limit points.

OhMyMarkov
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Hello everyone!

I'm trying to prove that the closure of $A = [-\infty,0)$ is $[-\infty,0]$. So far, I have proved that all points in $[-\infty, 0)$ are limit points of A, then I have proved that $\sup A = 0$, so it is in the closure, so $[-\infty, 0]$ subsets the closure.

But how do I know that it is equal to the closure?
 
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OhMyMarkov said:
Hello everyone!

I'm trying to prove that the closure of $A = [-\infty,0)$ is $[-\infty,0]$. So far, I have proved that all points in $[-\infty, 0)$ are limit points of A, then I have proved that $\sup A = 0$, so it is in the closure, so $[-\infty, 0]$ subsets the closure.

But how do I know that it is equal to the closure?

you have to prove that the closure is subset of $[-\infty, 0]$
In general if want to prove that
$A = B $
we have to prove the subset in both sides
A subset of B
B subset of A
how ? the easiest way ( usually ) let x in A prove it is in B this give A subset of B
and let x in B prove it is in A this give B subset of A
x arbitrary element
 

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