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QUESTION: $f(x,y)=y-\displaystyle\frac{1}{x^2}$
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.

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For $x<0$ the function is defined. Why the restriction $x>0$ when you are talking about its domain? –  drhab Aug 16 at 8:56
    
the question just asked to consider x>0 or am I confusing it? –  erin Aug 16 at 8:59
    
$D$ was referred to as the domain of $f$ and I do not see restrictions on it. $S$ was defined on base of $D$ with extra conditions among which $x>0$. –  drhab Aug 16 at 9:08

3 Answers 3

up vote 3 down vote accepted

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.

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$P_n \in S$ if $n > 1$ –  Darth Geek Aug 16 at 9:14
    
Thanks so much for all your help –  erin Aug 16 at 9:15
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@DarthGeek Thanks I have corrected this. –  Martin Sleziak Aug 16 at 9:16
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@erin I have changed notation to $g(x)$ and $h(x)$. (It was my mistake - I should not have used $f$, since another function is already denoted by $f$.) I have suggested $h(x)=\frac1{x^2}-x$ because $(x,y)\in S$ $\leftrightarrow$ $x>0$ and $y>h(x)$. –  Martin Sleziak Aug 16 at 9:25
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@erin They don't have to be the same. I have simply used the same values, because from the picture it is obvious that one of them is in $S$ and the other isn't. You can as well take $(1,1+\varepsilon)$ and $(1,1-\varepsilon)$. (This might be even better when doing the computation showing whether the point is in $S$.) I should have also written that $\varepsilon>0$, but this is probably clear from the context. –  Martin Sleziak Aug 16 at 9:33

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.

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Thank you for your help –  erin Aug 16 at 9:40

An easy way to see this is is:

The function which maps the set $D$ to the set $S$ is continuous, as such all attributes of the (topological) set $D$ are also attributes of the (topological) set $S$.

One can also find the attributes of $S$ directly, but lets use $D$.

$D$ is open, thus $S$ is also open.

$D$ is not bounded $S$ is also not bounded

$D$ is not compact $S$ is also not compact

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Sorry I only just started learning about Topology. –  erin Aug 16 at 9:03
    
What do you mean by: The function which maps the set D to the set S is continuous, as such all attributes of the (topological) set D are also attributes of the (topological) set S. –  erin Aug 16 at 9:03
    
@erin, no problem me too, however my answer is not complete, but gives a possible path to follow –  Nikos M. Aug 16 at 9:03
    
@erin the function being continous retains topological properties of the domain to the image set –  Nikos M. Aug 16 at 9:04
    
also in the answers the teacher has provided he said that it was Open, not closed, not bounded and not compact. I am really confused now –  erin Aug 16 at 9:04

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