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.
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Are you familiar with definition of continuous in topological spaces? – drhab Jul 07 '18 at 06:55
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not yet drhab :) – TheLast Cipher Jul 07 '18 at 07:00
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1To inform yourself you could take a look here. In fact the definition involving topological spaces is the "bottom" under all other definitions of continuity. – drhab Jul 07 '18 at 07:04
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Every function defined on the set of integers (endowed with the discrete Topology) is continuous.
Tsemo Aristide
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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
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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
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this makes me think when to use and when not to use visual intuitions. – TheLast Cipher Jul 07 '18 at 07:06