# Thread: is this set closed?

1. ## is this set closed?

this is a review problem that I am not sure about.

is the set {(x,y)∈R^(2 ) ∶ 3x^2+ 5y^2=1} a closed set?

my initial thought is yes however I think I am missing something important.

I know that this is an ellipse but it is not in standard form. I had mathlab graph it and it complained that there
complex numubers that it was going to ignore, so I suspect that while the graph appeared to be a solid curve it is actually
full of spaces between the real solutions.

any thoughts?

2. ## Re: is this set closed?

Do you mean a closed set in the usual topology on $R^2$? If so, is its complement open? That is, if you take a point (x, y) NOT satisfying that equation, must there be a (possibly very small) neighborhood about it that is also not in the set.

If (x,y) is not on the ellipse, draw a line from (x,y) perpendicular to the ellipse. That is the shortest distance from the point to the ellipse. What about a disk centered on (x, y) with radius half that distance?

I don't use mathlab but if you asked it to graph $y= \frac{1}{5}\sqrt{1- 3x^2}$ and $y= -\frac{1}{5}\sqrt{1- 3x^2}$ then it was "complaining" about x values larger than $\sqrt{1/3}$ and x values less than $-\sqrt{1/3}$. That is, outside the ellipse.

Yes, this is an ellipse- its "standard form" would be $\frac{x^2}{\left(\frac{1}{\sqrt{3}}\right)^2}+ \frac{y^2}{\left(\frac{1}{\sqrt{5}}\right)^2}= 1$