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Math Help - is this set closed?

  1. #1
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    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?
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  2. #2
    MHF Contributor

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    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
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