Question Solved1 Answer The set A is a triangle with dotted line which means it is a open set, and I need graph. 3. Let A = {(x1, x2) E R² | x1, x2 > 0, x1 + x2 < 1}. Show that every point T E A is an interior point of A. Use the definition of interior point. Hint: you need to find a radius r > 0, which will depend on the point ž, such that D,(ã) CA.
The
set A is a triangle with dotted line which means it is a open set,
and I need graph.
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Solution:-Given that,A={(x_(1),x_(2))inR^(2)∣x_(1),x_(2) > 0,x_(1)+x_(2) < 1}.let vec(x)=(x_(0),y_(0))in A=>x_(0)+y_(0) < 1 and x_(0) > 0y_(0)0 claim, vec(x) is an interior point of A.let L be the straight line x+y=1let d be the distare between bar(x) ard x+y=1Then d=|(x_(0)+y_(0)-1)/(sqrt(1+1))|=(1-x_(0)-y_(0))/(sqrt2) > 0.=>1-(x_(0)+y_(0))=sqrt(2d)". "choose r=d//2 then d > 0=>r > 0.claim Dr( vec(x))sub A.let vec(y)=(x_(1),y_(1))inD_(r)( vec(x))=>(x_(1)-x_(0))^(2)+(y,-y_(0))^(2)Now x_(1)+y_(1)-1=(x_(1)-x_(0)+x_(0))+(y_(1)-y_(0)+y_(0))=(4r^(2))/(2).{:[=(x_(0)+y_(0)-1)+(x_(1)-x_(0))+(y_(1)-y_(0))],[=-sqrt2d+(x_(1)-x_(0))+(y_(1)-y_ ... See the full answer