If $A \subset \mathbb{R}^{n}$ is not closed, show that there is a continuous function $f: A \to \mathbb{R}$ which is unbounded.

continuitymultivariable-calculusreal-analysis

If $A \subset \mathbb{R}^{n}$ is not closed, show that there is a continuous function $f: A \to \mathbb{R}$ which is unbounded. Hint: If $x \in \mathbb{R}^{n} \backslash A$, but $x \notin int(\mathbb{R}^{n} \backslash A)$, let $f(y) = \frac{1}{|y – x|}$.

I'm having trouble coming up with a function. The only thing that comes to my mind is that the hint they gave could be a function used to show the claimed result, but that would mean at least in my thought process that $y$ is approaching some value of $x$ but never arrives at it and as such the function will blow up. If this is the idea that should be conveyed then I'm struggling to convey it more formally.

Best Answer

The function is precisely the one given by the hint.

Suppose $A \subseteq \mathbb{R}^n$ is not closed. Therefore, there are limit points of $A$ which are not elements of $A$. Let $x$ be one such point, and define $f : A \to \mathbb{R}$ by $f(y) = \|x-y\|^{-1}$. Since $x \notin A$, this function is well-defined on $A$. Moreover, since $y \mapsto \|y\|$ is a continuous function on $\mathbb{R}^n$, $f$ is also a continuous function.

To see that $f$ is unbounded on $A$, fix a sequence $(x_n)_{n=1}^{\infty} \subset A$ such that $x_n \to x$. Then for every $\varepsilon > 0$ there is an $N \in \mathbb{N}$ such that $\|x-x_n\| < \varepsilon$ for each $n \geq N$. Let $M > 0$ be arbitrary. Then there is an $N \in \mathbb{N}$ such that $\|x-x_n\|< M^{-1}$ for each $n \geq N$. That is, $f(x_n) > M$ for each $n \geq N$. Since $M$ was arbitrary, we conclude that $f$ is unbounded on $A$.

To address the question in the comments, suppose that

  1. A set $U \subseteq \mathbb{R}^n$ is defined to be open if for each $x \in U$ there exists $\varepsilon > 0$ such that $y \in U$ for each $y$ satisfying $\|x-y\| < \varepsilon$.
  2. A set $F$ is defined to be closed if $\mathbb{R}^n \setminus F$ is open.
  3. The point $x$ is a limit point of the set $A$ if there exists a sequence $(x_n)_{n=1}^{\infty} \subseteq A$ with $x_n \ne x$ for each $n$ such that $x_n \to x$.

How do we know that a closed set contains its limit points?

Suppose $F$ is closed, and suppose $x$ is a limit point of $F$. Fix a sequence $(x_n)_{n=1}^{\infty} \subseteq A$ with $x_n \ne x$ for each $n$ such that $x_n \to x$. Then for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that $\|x-x_n\| < \varepsilon$ for each $n \geq N$.

Is $x \notin F$? For the sake of contradiction, assume so. Since $\mathbb{R}^n \setminus F$ is open, we can fix $\overline{\varepsilon} > 0$ such that for any $y$ satisfying $\|x - y\| < \overline{\varepsilon}$, we have $y \in \mathbb{R}^n \setminus F$. On the other hand, we can fix $\overline{N} \in \mathbb{N}$ such that $\|x - x_n\| < \overline{\varepsilon}$ for each $n \geq \overline{N}$. But each of these $x_n$ is an element of $F$. This is a contradiction.

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