Why does the Hausdorff distance between $I=[0,1]$ and $B_n = \{0, \frac{1}{n}, \cdots, \frac{n-1}{n}, 1\}$ tend to zero?

Question

I have a proof of this; I understand the implications of all the steps, but I am struggling to understand why a particular step is true, namely the statement

$B_n \subset I$ so we can say:$$\ \sup_{b \in B_n} \inf_{a \in I} d(a,b) = 0$$.

Why can we say this?

Any help would be appreciated!

Answer 1

Let $b \in B_n$. Then also $b \in [0,1]$, so $\inf_{a \in I} d(a,b) = d(b,b) = 0$.

Answer 2

Let us first clearly state the definition of the Hausdorff distance. For ease of reference, I will refer to the definition on Wikepedia which says, given a metric $d$ on a metric space and compact sets $X$ and $Y$,

$$d_H(X,Y) = \max\left(\sup_{x\in X}\inf_{y\in Y}d(x,y),\sup_{y\in Y}\inf_{x\in X}d(y,x)\right).$$

In the question at hand, $d$ is simply the standard metric on $\mathbb R$: $d(x,y)=|x-y|$. We are asked to show why $d_H(I,B_n)\to0$ as $n\to\infty$ where $I$ is the unit interval and $$B_n = \{0,1/n,2/n,\ldots,(n-1)/n,1\}.$$ In fact, we'll show that $d_H(I,B_n)=1/(2n)$.

We first claim that $$\sup_{y\in B_n}\inf_{x\in I}d(x,y) = 0.$$ This is a simple consequence of the fact that each $y\in B_n$ is also an element of $I$. Thus, $$\inf_{x\in I}d(x,y) = 0 \text{ for all } y\in B_n.$$ Next, we claim that $$\sup_{x\in I}\inf_{y\in B_n}d(x,y) = \frac{1}{2n}.$$ We can see this as follows: for every $n$, $i/n+1/(2n)$ is an element of $I$ whose distance from $B_n$ is $1/(2n)$. A look at a simple picture shows there are no points of $I$ that are farther away:

Of course, the actual Hausdorff distance is the larger of these two, namely $1/(2n)$ and $1/(2n)\to0$ as $n\to\infty$ to answer the question as currently asked.