MHB Limit Points of Unbounded Interval

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
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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