anonymous
  • anonymous
help here
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
yes
anonymous
  • anonymous
Let X be a topological space, Let \[{x_{n}} \] be a sequence of elements in X. Then\[ x_{n} \] is said to converge to x\[\epsilon \] X if \[\forall \] nbds U of x, there exists N\[\epsilon \] N such that \[\forall \] n\[\geslant \] N, \[x_{n} \]\[\epsilon \] U
anonymous
  • anonymous
then one of these conditions does not hold

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
A) \[x_{n} \rightarrow x , as.. n\rightarrow N \]
anonymous
  • anonymous
ya this is not in my range of knowledge, that looks like 20 years of college
zzr0ck3r
  • zzr0ck3r
The option you showed is true.

Looking for something else?

Not the answer you are looking for? Search for more explanations.