analysiscompactnesscontinuityreal-analysis
I have been struggling to show that there does not exist a surjective continuous map from $\mathbb{R}^2$ into $\mathbb{R}$ such that the inverse image of any compact set is compact.
I know that such a function would map compact sets to compact sets. Are there any suggestions for solving this problem?
Here's a non-trivial example: the closure of the topologist's sine curve with the ends joined up.
If you want an explicit representation, take $$ \{(x,y) : 0< x\leq 1, y = \sin(1/x) \} \bigcup \{(0,y): -1\leq y\leq 1\} \bigcup \{(x,0): -1\leq x\leq 0\} \bigcup \{(-1,y): -2\leq y\leq 0 \} \bigcup\{(x,-2): -1\leq x \leq 1\} \bigcup \{(1,y): -2\leq y\leq \sin(1) \}. $$
Take the identity function on set $X$ where the domain is equipped with discrete topology and the codomain with indiscrete topology.
That is a continuous bijection.
If $X$ has exactly $2$ elements then $X$ is compact but the singletons (which are open and closed) will be sent to sets that are not open and are not closed.
So the function is not open and is not closed.
Best Answer
Hint: assume $K=\{x:f(x)=0\}$ is compact. Take $R>0$ such that $K\subseteq B_R=\{x:\|x\|\le R\}$. Then $B_R$ is also compact and so $f[B_R]$ is bounded, say contained in $[-N,N]$ for some~$N$. Take $a$ and $b$ with $f(a)=-N-1$ and $f(b)=N+1$. Take a curve $C$ outside $B_R$ that connects $a$ and $b$ and show that there is a point on $C$ that maps to $0$.