How to find support of functions
Your solutions to (1) and (2) are correct.
In (3), the function $f$ is nonzero at least on $\mathbb Q^2$ (in fact on the bigger set $\mathbb Q\times\mathbb R\cup \mathbb R\cup \mathbb Q$). Since the rationals are dense in the reals, the closure of $\mathbb Q^2$ is already all of $\mathbb R^2$.
For (4), it may be advisable to sketch the area in question. We are given that $f$ is zero when $x+y\ge 1$; this is (the outside of) a diamond shaped figure. The expression $(xy)x+y(x+y)xy$ is zero if $xy$ and $x+y$ have the same sign, that is either $x+y\le 0$ and $x\le y$ (this is a rotated quadrant of the plane) or $x+y\ge 0 $ and $x\ge y$ (another rotated quadrant). On the other hand, if $xy$ and $x+y$ have opposite signs, then $$(xy)x+y(x+y)xy = \pm((xy)(x+y)+(x+y)(xy)) =\pm2(x^2y^2),$$ which is zero only on the diagonals (we already had that and it doesn't matter anyway when taking the closure). Now look at your sketch. It should look like two small diamonds.
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Siddhant Trivedi
Updated on October 11, 2020Comments

Siddhant Trivedi about 3 years
$\textbf{Support}$:$f$ is real valued function with domain $E^n$ the support of $f$ is the smallest closed set $K$ such that $f(x)=0$ for all $x$ is not in $K$
Find the support
$(1) f(x)=xx$
$\displaystyle(2) f(x,y)=\frac{x}{e^{x^2+y^2}} $
$(3) f(x,y)=1$ if either x or y is a rational number,$f(x,y)=0$ if both are irrationals.
$(4)f(x,y)=(xy)x+y(x+y)xy$ if $x+y<1$ otherwise $f(x,y)=0$ if $x+y\geq1$
I tried the first one$(1)$ it that function assumes nonzero values in the interval$(\infty,0)$and the smallest closed set containing this interval is $(\infty,0]$
In the second example,$f(x,y)=0$ iff $x=0$ therefore the set in which function assumes nonzero values is $E^2\{0\}$ the smallest closed set is $E^2$
please help me to solve $(3)$ and $(4)$.thanks in advance.

Siddhant Trivedi about 10 yearswhat happened at the boundary of diamond?i'm little bit confused at the boundary it may be zero.and under that diamond function is non zero that is open set.smallest close is $x+y\leq 1$ am i right?