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Define the function $f:\Bbb{Z}\rightarrow\Bbb{R}$ such that $f(z)=c$ and $c$ is any real number. Since $\forall \epsilon>0$, $\vert f(a)-f(b)\vert=\vert c-c\vert=0<\epsilon$ whenever $0<\vert a-b\vert<\delta$, for some $\delta >0$. Then $f$ must be continuous...? But the graph of this function have "breaks" in it. I'm trying to reconcile the graphical representation of continuity and its formal definition. Sorry if this is such a stupid question.

2 Answers2

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Every function defined on the set of integers (endowed with the discrete Topology) is continuous.

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If $X$ and $Y$ are topological spaces then every function $f:X\to Y$ that is constant is continuous.

This because every preimage under $f$ will be an element of $\{\varnothing,X\}$ and both sets are open by definition.

drhab
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  • Are you using a topological definition when you said "This because every preimage under $f$ will be an element of ${\varnothing,X}$ and both sets are open by definition."? I haven't taken topology yet but our lectures are heading that direction. – TheLast Cipher Jul 07 '18 at 06:58
  • Yes. If $X,Y$ are topological spaces equipped with topologies $\tau_X$ and $\tau_Y$ then by definition a function $f:X\to Y$ is continuous if $f^{-1}(U)\in\tau_X$ for every $U\in\tau_Y$. That is the definition I am using. – drhab Jul 07 '18 at 07:02
  • this makes me think when to use and when not to use visual intuitions. – TheLast Cipher Jul 07 '18 at 07:06