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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?

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

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

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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.

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