Suppose $X$ is a connected space, and $f:X\rightarrow \mathbb{R}$ a continuous real valued function. Is it true that $\{(x,y)\in X\times\mathbb{R}\mid f(x)>y\}$ is connected?


closed as off-topic by user223391, zz20s, JMP, Claude Leibovici, eranreches Dec 27 '17 at 12:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, zz20s, JMP, Claude Leibovici, eranreches
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ @ajotatxe if $f(x)-y$ is a function, I couldn't think out an example for that $\endgroup$ – 89085731 Dec 9 '17 at 19:33
  • $\begingroup$ Yes, I have deleted my comment for something. $\endgroup$ – ajotatxe Dec 9 '17 at 19:34
  • $\begingroup$ What did you try? Did you try the equality case $f(x)=y$? Do you already know that the product of connected spaces is connected? $\endgroup$ – Moishe Kohan Dec 9 '17 at 20:15

Yes. It is the image of the connected space $X\times \mathbb R_{>0}$ under the continuous map $$(x, t)\mapsto (x, f(x)-t)$$


Fix a $x_0\in X$. Consider these subsets of $X\times\mathbb R$:

  • $\{x_0\}\times(0,+\infty)$;
  • $\left\{\bigl(x,f(x)+k\bigr)\right\}$ ($k>0$).

They are all connected. So, for each $k>0$ the set$$G_k=\{x_0\}\times(0,+\infty)\cup\left\{\bigl(x,f(x)+k\bigr)\right\},$$since it's the union of two connected sets with a common point, which is $(x_0,k)$. But$$\bigl\{(x,y)\in X\times\mathbb{R}\,|\,f(x)>y\bigr\}=\bigcup_{k>0}G_k.$$Since each $G_k$ is connected and $\bigcap_{k>0}G_k=\{x_0\}\times(0,+\infty)$, which is not empty, $\bigl\{(x,y)\in X\times\mathbb{R}\,|\,f(x)>y\bigr\}$ is connected.


Not the answer you're looking for? Browse other questions tagged or ask your own question.