2

Does there exist a metric $d$ on $\mathbb R$ such that the map $f:(\mathbb R,d) \to (\mathbb R,d)$ defined as $f(x)=-x$ is not continuous?

Martin Sleziak
  • 56,060
  • 3
    There's a number of ways to do this - if I gave the hint that one such way would be to note that $f^{-1}((0,1))=(-1,0)$ and then make $(0,1)$ an open set in your metric space, but make $(-1,0)$ not open, could you come up with an example of a metric where this was satisfied? (If you come up with something, writing an answer to your own question is a good thing to do). Of course, this trick works with any $A$ and $B$ with $f^{-1}(A)=B$ if you can make $A$ open and $B$ not open. (This works well if $A$ and $B$ are singletons) – Milo Brandt Jun 14 '16 at 04:46

2 Answers2

0

Sure. Let $d |_{[1, + \infty)^{2}}, d |_{(- \infty, 1]^{2}} $ be equal to the standard metric, but if $x < 1 \leq y$ or $x \leq 1 < y$, then let $d(x, y) = |x| + |y| + 7$. You can check this is a metric. But you'll have that if $\epsilon < 7$, then we get weird behavior around $1$. Namely, we have discontinuity at $-1$.

Refer to the definition of continuity at $a$: For all $\epsilon > 0$ there exist $\delta > 0$ such that $d(x, a) < \delta$ implies $d(f(x), f(a)) < \epsilon$. Let $\epsilon_{0} = 1$. Then $B(f(-1), 1) = [1, 2) $. But if $\delta > 0$ is any positive real, then we may consider $x = \min \{ - 1 + \delta / 2, - 1 / 2 \}$ which is within distance $\delta$ of $- 1$, but $f(x) = \max \{ 1 - \delta / 2, 1 / 2 \}$, but $d(f(x), 1) \geq 7 > 1$. Thus $f$ cannot be continuous at $-1$.

AJY
  • 9,086
0

An easy metric construction is $d'(x,y)=d(g(x),g(y))$ where $g$ is a bijection and $d$ is another metric on the base set. In the new metric the open sets are the $g$-images of open sets in the original metric. So all we have to do is come up with a bijection that maps $f$ to a discontinuous function.

For example, let $g$ swap $0$ and $1$, that is $g(0)=1$, $g(1)=0$ and $g(x)=x$ for $x\ne0,1$. Then the $g$-image of $f$ is $f^g=g^{-1}\circ f\circ g$, which has values $f^g(0)=-1$, $f^g(1)=1$, $f^g(-1)=0$, and $f^g(x)=-x$ for other values of $x$. Clearly this function is not continuous around any of $\{-1,0,1\}$ in the standard metric, so $f$ is not continuous around the $g$-images of these points (which is also $\{-1,0,1\}$) in the new metric.

Mario Carneiro
  • 28,221