Determine whether Set is closed,whether it is open, whether it is bounded and whether it is compact

Question

Question:

Given the $$f(x,y)=y-\displaystyle\frac{1}{x^2}$$ consider the set $S = \{(x,y) \in D: x > 0,\ f(x,y) > 0\}$ where $D$ is the domain of $f$. Sketch the set $S$ in the plane. Determine whether $S$ is closed,whether it is open, whether it is bounded and whether it is compact. Explain carefully

---

Attempt at answer:

The domain will be $x>0$ and $y>\displaystyle\frac{1}{x^2}$

A function is closed if it contains all of its boundary points

I’m not sure if I am correct in saying but since it is not a strictly greater than and equal than it does not include all of the boundary points Also $(1,1)$ is a boundary point in $S$ but it is not an element of the set as $1=\displaystyle\frac{1}{1^2}$

A function is open if all points are interior points. Not to sure how to show this one?

A function is bounded if you can draw a ball of radius $k$ such that the entire set is contained This is not possible as the function is infinite so will always have points outside the ball

A function is compact if it is closed and bounded. Since it is neither closed nor bounded it is not compact.

Answer 1

Let me go through your attempted answer:

The domain will be $x>0$ and $y>\displaystyle\frac{1}{x^2}$

If you meant to say that $S=\{(x,y); x>0, y>\frac1{x^2}\}$, then this is correct.

If you meant $D$ by the word domain, then this is incorrect: $D$ is the set of all points for which $f(x,y)$ is defined.

A function is closed if it contains all of its boundary points

You meant to say that a set (not a *function) is closed if and only if it contains all boundary points. (But this is probably just a typo.)

Also $(1,1)$ is a boundary point in $S$ but it is not an element of the set as $1=\displaystyle\frac{1}{1^2}$

Yes, this is true. But maybe you should also explain why $(1,1)$ is a boundary point.

A function is open if all points are interior points. Not to sure how to show this one?

If you have $(x,y)\in S$ then $$y>\frac1{x^2}$$ and $x>0$.

Using continuity of the function $g(x)=1/x^2$ (or maybe using $h(x)=\frac1{x^2}-x$ would be even simpler), can you show that there exists $\varepsilon>0$ such that $$ |s-x|<\varepsilon, |t-y|<\varepsilon \qquad \implies \qquad s>\frac1{t^2}?$$ This basically says that points close enough to $(x,y)\in S$ are still in $S$. Showing this is sufficient to get that each $(x,y)\in S$ is an interior point.

A function is bounded if you can draw a ball of radius $k$ such that the entire set is contained

You should write set instead of function.

This is not possible as the function is infinite so will always have points outside the ball.

I get your point, but perhaps you should explain it more in detail. For example the points $P_n=(n,n)$ for $n>1$ all belong to $S$. Is it clear that for any ball $B(0,r)$ around the origin one of this points will be outside the given ball?

A function is compact if it is closed and bounded. Since it is neither closed nor bounded it is not compact.

Again set instead of function. Otherwise ok.

Answer 2

What you wrote is the domain of $f$ restricted to $S$. The actual domain of $f$ is $\mathbb{R}^2 - \{(0,y): y\in \mathbb{R}\}$.

As you said, $(1,1)$ is a boundary point of $S$ that doesn't belong in $S$ so $S$ is not closed.

To show that all it's points are interior, consider this:

Let $(x_0,y_0)\in S$ then $x_0 > 0$ and $x_0^2y_0 > 1$. Let $d = \min\lbrace\sqrt{(x_0-x)^2+(y_0-y)^2}:x^2y=1\rbrace$ Consider $k = \min\{x_0,d\}$

Then $B\left((x_0,y_0),k\right)\subseteq S$. Therefore all points in $S$ are interior.

Your proof that $S$ is not bounded seems a bit dodgy. Perhaps this helps. $c:=(1,2)\in S$. So if $S$ is bounded there is a $k\in R$ so that $S \subseteq B(c,k)$ But $(2k+1,2)\in S$ while $(2k+1,2)\notin B(c,k)$. Therefore, $S$ is not bounded.

Indeed, if $S$ is not closed or not bounded it is not compact.