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Question: 1-28. If $A \subset R^{n}$ is not closed, show that there is a continuous function $f: A \rightarrow R$ which is unbounded.

Hint : If $x \in R^{n} - A$ but $x \notin int(R^{n} - A)$, let $f(y) = \frac{1}{ \mid y-x \mid}$.

What I got so far:

Given $f(y) = \frac{1}{\mid y-x \mid}$, choose $p$ to be the point that's in the not closed set.

So, for $x \in A$ and $p \notin A$, our function becomes $f(x) = \frac{1}{\mid x-p \mid}$

By the continuous definition, for $ \epsilon > 0$ there exists a $\delta >0$ such that $\mid x-y \mid < \delta$ implies $\mid f(x)-f(y) \mid < \epsilon$.

$\mid \frac{1}{\mid x-p \mid} - \frac{1}{\mid y-p \mid} \mid < \epsilon$  

$\mid \frac{\mid y-p \mid}{\mid y-p \mid} \cdot \frac{1}{\mid x-p \mid} - \frac{1}{\mid y-p \mid} \cdot \frac{\mid x-p \mid}{\mid x-p \mid} \mid < \epsilon$

$\mid \frac{\mid y-p \mid -\mid x-p \mid }{\mid y-p \mid \mid x-p \mid } \mid < \epsilon$

$\mid \frac{\mid y-p-x+p \mid }{\mid y-p \mid \mid x-p \mid } \mid < \epsilon$

$\mid \frac{\mid y-x \mid }{\mid y-p \mid \mid x-p \mid } \mid < \epsilon$

I need to find two Deltas but I'm not sure how to do that. Does it have something to do with using

$\mid \mid x \mid - \mid y \mid \leq \mid x-y \mid$ ?

Or maybe I'm doing this backwards because if I take the negative out of the numerator, I'll end up with $-\mid -y+x \mid$.

Anik Bhowmick
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usukidoll
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  • Use a more general fact: if $g$ is continuous and nonzero, then $1/g$ is continuous as well. I'm sure this is proven in whatever book you're using. – David Bowman Sep 05 '18 at 02:32
  • I had a similar example done with x being continuous and if x isn't zero then $\frac{1}{x}$ is also continuous. In the example I saw $\mid x-a \mid < \delta \rightarrow \mid f(x)-f(a) \mid < \epsilon$. The end result was $\mid \frac{a-x}{ax} \mid$ for $ax \neq 0$ and it was bounded from below. But the thing is that supposedly I have to find the Deltas and that's confusing. There's supposed to be two of them. How does it go from whatever I type to the Deltas being $\frac{\mid a \mid}{2} and $\frac{ \mid a^2 \mid \epsilon}{2}$ – usukidoll Sep 05 '18 at 02:38
  • If only that part was better explained I can get the two Deltas I need for this problem. – usukidoll Sep 05 '18 at 02:40

2 Answers2

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Since $X \subseteq \mathbb{R}^n$ is not closed, there exists $a \in \mathbb{R}^n$ and a sequence $\{x_n\}$ in $X$ such that $\lim_{n\to\infty}x_n=a$ but $a \not\in X$. Without loss of generality, you can assume that $x_n$ is never equal to $a$ for any $n \in \mathbb{N}$.

Now consider $f(x) = \frac{1}{\|a-x\|}$. Since $\lim_{n\to\infty}x_n=a$, we know that:

$$\forall K \in \mathbb{N}, \exists M\in\mathbb{N}: n\geq M \implies \|a-x_n\|<\frac{1}{K}$$

This means that $$\forall K\in \mathbb{N}:|f(x_n)|=\frac{1}{\|a-x_n\|}>K$$

Hence, $f$ is not bounded.

To see that $f(x)$ is continuous in a punctured neighborhood of $a$, note that $f(x)$ is a rational function where the denominator is zero only at $x=a$. So, $f(x)$ is indeed continuous over $\mathbb{R}^n - \{a\}$. To follow an $\epsilon-\delta$ approach, you need to review and combine the following two proofs:

  1. $f(x) = \frac{1}{x}$ is continuous over $\mathbb{R}-\{0\}$
  2. If $g$ and $f$ are continuous at $x=c$, so is $f\circ g$.

Proving the second statement is straightforward and easy. I leave it to you. In our case, it suffices to show that $f(x) = \frac{1}{x}$ is continuous over $(0,\infty)$. We first observe that for a fixed value $y \in (0,+\infty)$, we have:

$$|f(x)-f(y)| = |\frac{1}{x}-\frac{1}{y}|=|\frac{y-x}{xy}|\leq\frac{|x-y|}{y}\cdot\frac{1}{x}$$ Now, our goal is to select $\delta$ such that it's easy to find an upper bound for $\frac{1}{x}$ when $x \in (y-\delta,y+\delta)$. Equivalently, we want to choose $\delta$ such that $x \in (y-\delta,y+\delta)$ is away from $0$. Choosing $\delta < \frac{y}{2}$ should do the job for us. Indeed, you can easily show that in this case we have $\frac{1}{x} < \frac{2}{y}$. This means that

$$|f(x)-f(y)| \leq \frac{|x-y|}{y}\cdot\frac{2}{y} \leq \frac{2\delta}{y^2}$$

Now choosing $\delta < \min\{\frac{y}{2},\frac{y^2}{2} \epsilon\}$ finishes the proof. Q.E.D.

stressed out
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  • Your username matches what I'm feeling right now lol. But I'm supposed to find the two deltas for this...how???? – usukidoll Sep 05 '18 at 02:42
  • @usukidoll Yeah. My username is what most math students feel at most times. That's why I chose it xD Why do you insist on using the epsilon-delta approach? Is there a particular reason for that? – stressed out Sep 05 '18 at 02:45
  • It has to be done using the epsilon-delta approach. – usukidoll Sep 05 '18 at 02:46
  • My book Calculus on Manifolds by Spivak has the continuous function with the delta and epsilon so I have to use that definition. I have two other books that also uses the epsilon-delta approach as well. What I am having a hard time with is finding the deltas. :/ – usukidoll Sep 05 '18 at 02:49
  • @usukidoll I don't see why 'it has to be done that way'. That's certainly not stated in the problem itself. (I have read Calculus on Manifolds by Spivak). But if you really insist on that, you can check any calculus book about why $f(x)=\frac{1}{g(x)}$ is continuous where $g(x)$ does not vanish. The proof will be the same. If my memory serves me well, this has been proved in Apostol's Calculus, Vol. 1. I can write down the proof for you if you want. – stressed out Sep 05 '18 at 02:51
  • Yes can you please write down the proof to that? I can follow along when I redo it on scratch paper. – usukidoll Sep 05 '18 at 02:52
  • @usukidoll see this post for example: https://math.stackexchange.com/questions/12462/is-fx-1-x-continuous-on-0-infty I will extend my post after dinner. – stressed out Sep 05 '18 at 02:57
  • @usukidoll I added an epsilon-delta proof for you. Let me know if you needed more help. – stressed out Sep 05 '18 at 04:02
  • How were you able to get the delta to be $\frac{y}{2}$ and then the second delta as $\frac{y^2}{2} \epsilon$? I recognized the $(y-\delta)(y+\delta)$ as a neighborhood, so x is in that neighborhood so do we need to choose a Delta that's inside there? – usukidoll Sep 05 '18 at 04:08
  • @usukidoll we know that $x \in (y-\delta, y+\delta)$. Right? We want $x$ to be away from zero, which happens when $y-\delta>0$ because then $\frac{1}{x}$ will be bounded from the above by $\frac{1}{y-\delta}$. Do you agree? Well, a natural choice to make this happen is $\delta = \frac{y}{2}$ because then $x \in (\frac{y}{2},\frac{3y}{2})$ which is away from zero. For the second part, notice that when $x < \min{a,b}$, both of the inequalities $x < a$ and $x < b$ hold true. Now just plugin delta in the last inequality and you see that $|f(x) - f(y)| < \epsilon$. – stressed out Sep 05 '18 at 04:12
  • When I fix $\delta = \frac{y}{2}$ I did get that interval so x has to be away from zero but the delta has to be as small as possible...as in less than the epsilon. So for my problem I have to choose a Delta small enough to be less than the epsilon and away from zero. So the neighborhood may be like $ x \in ( \mid y-p \mid - \delta, \mid y-p \mid + \delta)$. – usukidoll Sep 05 '18 at 04:19
  • @usukidoll delta doesn't have to be smaller than epsilon. $|f(x) - f(y)|$ has to be smaller than epsilon. And if you choose delta smaller than $\frac{y}{2}$ and $\frac{y^2 \epsilon}{2}$, then the last inequality tells you that $|f(x) - f(y)| < \epsilon$. Right? – stressed out Sep 05 '18 at 04:22
  • Right .... ...... – usukidoll Sep 05 '18 at 04:23
  • Let us continue this discussion in chat. – stressed out Sep 05 '18 at 04:24
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Supongamos que $A\subset\mathbf{R}^{n}$ no es cerrado. Entonces, existe un punto $x\in \mathbf{R}^{n}\setminus A$ que pertenece a la adherencia de $A$. Esto significa que para cualquier vecindad $V$ de $x$, existe algún punto $y\in V\cap A$.

Definamos ahora la función $f:A\to \mathbf{R}$ como $f(y) = \frac{1}{\lvert y - x\rvert}$. Notemos que esta función está bien definida ya que $x\notin A$.

Por otro lado, veamos que $f$ es continua. Sea $y_{0}\in A$ y $\epsilon > 0$. Debemos encontrar $\delta > 0$ tal que $\lvert y - y_{0}\rvert < \delta$ implica que $\lvert f(y) - f(y_{0})\rvert < \epsilon$.

Para ello, tomamos $\delta = \frac{1}{\epsilon}$. Entonces, si $\lvert y - y_{0}\rvert < \delta$, se tiene que:

$$\lvert f(y) - f(y_{0})\rvert = \left\lvert\frac{1}{\lvert y - x\rvert} - \frac{1}{\lvert y_{0} - x\rvert}\right\rvert \leq \frac{\lvert \lvert y - x\rvert - \lvert y_{0} - x\rvert\rvert}{\lvert y - x\rvert \lvert y_{0} - x\rvert}$$

Pero como $y_{0}\in A$, se tiene que $\lvert y_{0} - x\rvert > 0$, y por lo tanto:

$$\frac{\lvert \lvert y - x\rvert - \lvert y_{0} - x\rvert\rvert}{\lvert y - x\rvert \lvert y_{0} - x\rvert} \leq \frac{\lvert y - y_{0}\rvert}{\lvert y - x\rvert \lvert y_{0} - x\rvert} < \frac{\delta}{\delta^{2}} = \epsilon$$

Por lo tanto, $f$ es continua en $A$. Sin embargo, notemos que $f$ es no acotada ya que para cualquier $y\in A$, se tiene que $\lvert f(y)\rvert > \lvert f(x)\rvert = \frac{1}{\lvert x - y\rvert}$. Como $x$ pertenece a la adherencia de $A$, existen puntos de $A$ arbitrariamente cercanos a $x$, y por lo tanto $f$ no está acotada en $A$.

Susi
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