Prove that there is no continuous bijection $f:\mathbb Q\to K$ of $\mathbb Q$ on a compact metric space $K$ Help
Asked
Active
Viewed 84 times
-1
-
2What is $Q?{}{}$ – Thomas Andrews May 12 '21 at 22:41
-
1Q is Rational, sorry – Itzel Najera May 12 '21 at 22:43
-
3What have you tried? – paul blart math cop May 12 '21 at 22:46
-
2Consider and enumeration $x_1,x_2,x_3,...$ of $\mathbb{Q}$ and the sets $A_n=f({x_1,x_2,...,x_n})\subset K$. These sets must be nowhere dense. Then use Baire's theorem. – plop May 12 '21 at 22:54
-
3One way is to show that a countable compact metric space must have an isolated point and then use the continuity of $f$ to get a contradiction; the answers to this question may be helpful. – Brian M. Scott May 12 '21 at 22:57
-
Help me plis... – Itzel Najera May 13 '21 at 02:56
1 Answers
1
Let $q_n \rightarrow r$ where $q_n$ are rational and $r$ is irrational. Since $f$ is continuous $f(q_n)$ is cauchy. Since $K$ is compact $f(q_n)$ has a subsequency converging within $K$, Let wlog $f(q_n) \rightarrow z$, $z \in K$. Since $f$ is a bijection, $z=f(c)$. Hence $f(q_n) \rightarrow f(c)$ with $c \in \mathbb{Q}$. So if inverse of $f$ is continuous then $f^{-1}(f(q_n)) \rightarrow f^{-1}(f(c))$. hence $q_n \rightarrow c$ but $c$ is a rational number. hence a contradiction. I am assuming that inverse is also continuous.
Balaji sb
- 4,357