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Context:

I am taking a course in analysis at the university. Many problems on $\mathbb{R}$ are easy under Euclidean distance, but I find them difficult under a general metric. So I wonder if an open interval must be open under any metrics.

Consider the metric spaces in the form $(\mathbb{R}, d)$, where $d: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ may not be Euclidean.

Is there any metric space $(\mathbb{R},d)$ such that there exists some open interval $(a,b) \subseteq \mathbb{R}$ which is not open in $\mathbb{R}$?

My attempt:

Suppose $(a,b)$ is not open. Then there exists $x \in (a,b)$ such that for any $\varepsilon > 0$, $B(x,\varepsilon) \not \subseteq (a,b)$.

For such a point $x$, for $0 < \varepsilon < x - a$, consider point $y \in B(x,d(x,a + \varepsilon)) \setminus (a,b)$. Then we have $d(x,y) < d(x,a + \varepsilon)$ and $(y \leq a \text{ or } y \geq b)$. Suppose such a point $y \leq a$. Then $$ d(y, a + \varepsilon) \leq d(x, y) + d(x, a + \varepsilon) < 2d(x, a + \varepsilon). $$

But this seems to fail.

Thanks in advance.

Another User
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Timothy Fuh
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2 Answers2

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Consider the function$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x=-1\\-1&\text{ if }x=1\\x&\text{ otherwise.}\end{cases}\end{array}$$and the distance $d(x,y)=|f(x)-f(y)|$. Then $(0,2)$ is not an open set in $(\Bbb R,d)$ because, in $(\Bbb R,d)$, every open set to which $1$ belongs has to contain a set of the form $(-1-\varepsilon,-1+\varepsilon)\setminus\{1\}$, for some $\varepsilon>0$.

Another User
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2

Yes. Take a bijection $f : \{0,1\}^{\mathbb{N}} \to \mathbb{R}$. Endow the set $\{0,1\}^{\mathbb{N}}$ with the usual product metric, this is equivalent to the Cantor set. Use $f$ to define a metric on $\mathbb{R}$, in such a way that $f$ becomes an isometry. That is, $d(x,y) := d(f^{-1}(x),f^{-1}(y))$ for $x,y \in \mathbb{R}$. This generates a topology on the set of real numbers which is compact. If for this topology every interval is open, then $(0,1) = \bigcup_{0 < p < 1} (p,1)$ is an open covering that doesn't have a finite subcover. So the topology is not compact, and we have a contradiction.

Martin Brandenburg
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