Subset of continuous functions from [0,1] to R closed for the Hausdorff metric

Question

For the past few days I've been studying metric spaces and I was just making some exercices since I have an exam on it in a few days. I was struggling a bit with the following problem: let $G$ be the set of all graphs of continuous function from $[0,1]$ to $\mathbb{R}$. I've already proved that this set $G$ is a closed and bunded set of $\mathbb{R}^2$ for the standard metric. Thereforde we can see $G$ as being a subset of $F$ where $F$ is the set of all non-empty, closed and bounded sets of $\mathbb{R}^2$. The question now is if $G$ is closed for the Hausdorff metric. I approached this problem by chosing a random sequence of graphs in $G$ which is convergent to some set in $F$ and i then wanted to proof that this limit set is an element of $G$. I think this is the easiest approach however I can't seem to find a good proof of it since I always find exercices on the Haussdorf metric difficult. For the rest I know that $G$ is not open. Does anyone have a tip for proving that $G$ is closed, as always any help would be greatly appreciated :))

Answer 1

The sets of graphs of continuous functions is not closed with the Hausdorff metric. We'll show a counter example for a sequence of continuous functions in $[-1,1]$ that I trust it's illustrative enough to reproduce in $[0,1]$ (Admittedly I noticed the domain was $[0,1]$ after I finished writing this).

Consider the sequence of functions $$f_{n}(x)=\begin{cases} -1 & x \leq \frac{-1}{n} \\ nx & |x| \leq \frac{1}{n} \\ 1 & x \geq \frac{1}{n} \\ \end{cases} $$

And $G_{n}$ their graphs. Now consider the set $$K =[-1,0]\times \{0\} \cup \{0\} \times [-1,1] \cup [0,1] \times \{0\}$$

By drawing these sets it should be intuitive that $G_{n} \xrightarrow{H}K $, lets prove this.

For a set $A$ we note $A^{\varepsilon}=\{ x : d(x,A) < \varepsilon\}$. To prove convergence in the Hausdorff metric, we have to show that for every $\varepsilon >0$ there is $N$ large enough so that $K\subseteq G_{n}^{\varepsilon}$ and $G_{n}\subseteq K^{\varepsilon}$ for every $n > N$.

Take $\frac{1}{n} < \varepsilon$. For $(x,f_{n}(x))\in G_{n}$, if $|x| \leq \frac{1}{n}$ clearly $(x,f_{n}(x)) \in K$. If $|x| \geq \frac{1}{n}$ we have to find an element $y\in K$ so that $d((x,nx),y)< \varepsilon$. We take $(0,nx)$ so $d((x,nx),(0,nx))=|x|\leq \frac{1}{n}<\varepsilon$.

We see $G_{n} \subseteq K^{\varepsilon}$ for $n$ large enough so that $\frac{1}{n} < \varepsilon$.

For the other contention, we first consider the case $(x,-1)\in K$ with $-1 \leq x\leq 0$.

If $|x| \geq \frac{1}{n}$, clearly $(x,-1)\in G_{n}$. Otherwise notice $(\frac{-1}{n},-1)\in G_{n}$, so $$d((x,-1),(\frac{-1}{n},-1))=\frac{1}{n} + x\leq \frac{1}{n} < \varepsilon$$

The case $(x,1) \in K$ is similar. For the case $(0,y)\in K$ with $-1 \leq y \leq 1$, we have $(\frac{y}{n},y)\in G_{n}$, so $$d((0,y),(\frac{y}{n},y))=\frac{|y|}{n}<\varepsilon$$

And again we have $K \subseteq G_{n}^{\varepsilon}$ for large enough $n$.