1. ## IS it complete?

Is the set D={(x,y)∈R²:x²+y²<1} complete?

how?
actually i am in dark how to do it.

2. Originally Posted by sorv1986
Is the set D={(x,y)∈R²:x²+y²<1} complete?

how?
actually i am in dark how to do it.

There is a common theorem which says that for subspaces of a complete space completeness and closedness are synonomous. To be more explicit, if $\displaystyle D$ were complete the sequence $\displaystyle \displaystyle 1-\frac{1}{n}$ (I'm thinking in $\displaystyle a+bi$ notation) would converge to some point of $\displaystyle D$. Does it?

3. Originally Posted by Drexel28
There is a common theorem which says that for subspaces of a complete space completeness and closedness are synonomous. To be more explicit, if $\displaystyle D$ were complete the sequence $\displaystyle \displaystyle 1-\frac{1}{n}$ (I'm thinking in $\displaystyle a+bi$ notation) would converge to some point of $\displaystyle D$. Does it?
a lot of thanx. i got the concept what you wanna said.But (1,-1/n) belongs to that D? hey!i am a fan of yours.

4. Originally Posted by sorv1986
a lot of thanx. i got the concept what you wanna said.But (1,-1/n) belongs to that D? hey!i am a fan of yours.
Haha, thanks. I'm not quite sure what you mean. The sequence I mentioned as actually $\displaystyle \left(1-\frac{1}{n},0\right)$ which is Cauchy in $\displaystyle D$ yet converges to $\displaystyle (1,0)\notin D$.

5. Originally Posted by Drexel28
Haha, thanks. I'm not quite sure what you mean. The sequence I mentioned as actually $\displaystyle \left(1-\frac{1}{n},0\right)$ which is Cauchy in $\displaystyle D$ yet converges to $\displaystyle (1,0)\notin D$.
got it. Then any subset of R² which is closed, is complete and which is not closed, is not complete. is it right?

6. Originally Posted by sorv1986
got it. Then any subset of R² which is closed, is complete and which is not closed, is not complete. is it right?
Right--it holds more generally for any complete metric space.