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$x\in X$

Show that $\{y \in X \mid d(y,x) > r\}$ is open for every $r \in \mathbb{R}$

My first thought was that what the distance between y and x, it is a closed set. Apparently, it is not. I'm trying to figure out, how to prove this. I haven't gone far because I don't know the intuition. Can anyone help me get started?

2 Answers 2

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Let $U$ be the set in question. Then $B(y,{1 \over 2} (d(x,y)-r)) \subset U$, hence $U$ is open.

Another way to look at it is to notice that $f(y) = d(y,x)$ is continuous and $U = f^{-1} ((r,\infty))$. Since $f$ is continuous, the inverse image of an open set is open.

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Define $O = \{y : d(y,x) > r\}$. To see that $O$ is open we will show that every $p \in O$ is an interior point of $O$.

Let $p$ be such that $d(x,p) > r$, for some fixed $r$.

Then our "wiggle room" for $p$ is $\delta := d(x,p) - r > 0$.

Then $B(p, \delta) \subset O$, so $p$ is an interior point of $O$: let $q \in B(p, \delta)$.

Suppose that $d(q,x) \le r$, then we would have by the triangle inequality:

$$d(p,x) \le d(p,q) + d(q, x) < \delta + r = d(x,p)$$

which is a blatant contradiction. The last equality follows from the definition of $\delta$ and the second $<$ from $q \in B(p, \delta)$.

So $d(x,q) \le r$ is false, hence $q \in O$. As $q \in B(p, \delta)$ was arbitrary, we are done: all points of $O$ are interior points.

Note that this proof works in any metric space.

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    Thank you very much. I have used your help, and tried to write like this: – 2017-02-03
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    If we can show, that for $\forall y \in A$ $\exists \delta>0: B_{\delta}(y) \subseteq A$ we have proven, that the subset $A \subseteq X$ is open\\ We choose $\delta=d(x,y)-r>0$ \emph{(which is greater than 0 by the definition above.)} Let $q \in B_{\delta}(y)$ $\Rightarrow d(y,q)<\delta$\\ Now we assume that: $d(x,q) – 2017-02-03
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    Now we have contradiction: (The last inequality comes from: $d(y,q)<\delta$, since $q \in B_{\delta}(y)$.) $d(x,q) $q \in A$, is center of an open ball, which is contained in the subset A. – 2017-02-03
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    You need to assume $d(x,q) \le r$, not just $d(x,q) < r$, so get $q \in A$. The one strict inequality $d(q,y) < \delta$ is enough to get the contradiction ($(d(x,y) < d(x,y)$),( as $a < b$ and $c\le d$ implies $a+c < b+d$for reals $a,b,c,d$ ) $ q \notin A$ means $d(x,q) \le r$, not $<$). – 2017-02-04