Prove that $\frac{xy}{x^2+y^2}$ is not continuous

Question

Let $f:\mathbb R\times \mathbb R\to \mathbb R$ a function with two variables, prove that the function is not continuous $$f(x,y)=\begin{cases}\dfrac{xy}{x^2+y^2}, & \forall(x,y)\neq (0,0)\\ \\ 0, & (x,y)=(0,0)\end{cases}$$

Note: this question from topology course, not from calculus

Attempt:

Let $X=\mathbb R\times \mathbb R,\\ Y=\mathbb R\\ (x_0,y_0)=(0,0)\\ f\left(x_0,y_0\right)\in U\subseteq Y$

Now I should show that $\exists V\subseteq X$ such that $x\in V$ and $f(V)\subseteq U$

How can I prove that?

for homogeneous numerator and homogeneous denominator, it is enough to switch to polar coordinates — Will Jagy, May 23 '17 at 16:54
What do you say you have to prove?? That looks pretty weird...And how is that related to proving $;f;$ isn't continuous? — DonAntonio, May 23 '17 at 16:54
@Error404 I've no idea what's that to do with what you wrote. In order to prove continuity you must show the inverse image of any open set is open... — DonAntonio, May 23 '17 at 17:06
@DonAntonio Yes, this is the kind of proof that I'm looking for — Error 404, May 23 '17 at 17:29

score 3 · Answer 1 · answered May 23 '17 at 16:53

3

Hint. What happens along the lines $y=0$ and $y=x$?

answered May 23 '17 at 16:53

Robert Z

145,942

score 1 · Answer 2 · answered May 23 '17 at 16:54

1

If you let $x=0$ and take the limit as $y \to 0$, you get $0/y^2= 0$.

If you let $x=y$ and take the limit as $y \to 0$ you get $1/2$.

answered May 23 '17 at 16:54

gt6989b

54,422

score 1 · Answer 3 · answered May 23 '17 at 16:57

Consider $(x,y)\to(0,0)$ in various radial directions, say along $y=kx$.

Note that the limit will then depend on the direction of approach (i.e., on $k$) so that the limit does not exist. To exist, it must be independent of the manner of approach.

score 1 · Answer 4 · edited May 24 '17 at 05:28

1

If you're interested in proving the non-continuity of a function at a certain point, try to find two different paths on which to approach the point in question (i.e. take two different $\lim$) along the function graph which give two different values. Remember the definition of continuity. Hope this helps

edited May 24 '17 at 05:28

Vincenzo Tibullo

10,955

answered May 23 '17 at 16:59

garex434

39

score 1 · Accepted Answer · answered May 23 '17 at 19:36

To give a direct topological proof that the function is not continuous at the origin, suppose it were continuous and derive a contradiction. So, using the continuity assumption with $U = B_{1/3}(0)$, you would be able to find a neighborhood $V$ of $(0, 0)$ such that $V \subseteq f^{-1}(U)$. Since $V$ is a neighborhood, then for some $\delta > 0$, we have $B_\delta((0, 0)) \subseteq V$. Then we can conclude that whenever $\lVert (x, y) \rVert < \delta$, $|f(x, y)| < \frac{1}{3}$. However, note that for any $t \ne 0$, $f(t, t) = \frac{1}{2}$, and it is possible to choose $t$ such that $\lVert (t, t) \rVert < \delta$, which gives the desired contradiction.

(Or, for a slightly less direct proof, you can use the fact that if $f$ were continuous, then we would have that for any sequence $(a_n)_{n=1}^\infty \subseteq \mathbb{R}^2$, if $a_n \to (0, 0)$, then $f(a_n) \to f(0, 0) = 0$. However, then you can get a contradiction using the sequence $a_n = (\frac{1}{n}, \frac{1}{n})$.)

Vincenzo Tibullo · Answer 6 · 2017-05-24T04:43:04.847

If $f$ were continuous then the pre-image of every open set would be an open set, but also the pre-image of every closed set would be a closed set.

We can see that $$ f(x,y)=\frac12 \frac{2xy}{x^2+y^2}=\frac12\implies x^2+y^2=2xy\implies(x-y)^2=0\implies x=y $$ So the function assumes the value $1/2$ on all the points of the line $x=y$, and in no other points, but excluded $(0,0)$ (where it assumes the value $0$). So $$ f^{-1}(1/2)=\{(x,y)\in\mathbb{R}^2:x=y\text{ and }(x,y)\neq(0,0) \} $$ and $\{1/2\}$ is closed but $f^{-1}(1/2)$ is not.

Prove that $\frac{xy}{x^2+y^2}$ is not continuous

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