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I'm having some trouble understanding a step of the proof of the following theorem: if $f$ is continuous at $g(c)$ and $g$ is continuous at $c$, then $fog$ is continuous at $c$.

Proof:

Step 1: given that $f$ is continuous at $g(c)$, then there exists $\delta_1$ such that ${|f(t)-f(g(c))|}$ < $\epsilon$ when ${|t-g(c)|}$ < $\delta_1$. I have no problem with this step.

Step 2: given that $g$ is continuous at $c$, then there exists $\delta$ such that ${|g(x)-g(c)|}$ < $\delta_1$ when ${|x-c|}$ < $\delta$. I also understand this step.

Last step: what I don't understand is why it follows from steps 1 and 2 that if ${|x-c|}$ < $\delta$, then ${|f(g(x))-f(g(c))|}$ < $\epsilon$ .

Could you clarify?

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  • $\begingroup$ Has your question been answered? If yes, you should accept an answer. $\endgroup$ – Haris Gušić Jun 24 at 15:08
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If $|x-c| < \delta$ then $|g(x) - g(c)| < \delta_1$, by step 2.

If $|g(x) - g(c)| < \delta_1$, then $|f(g(x)) - f(g(c))| < \epsilon$, by step 1 applied with $t = g(x)$.

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  • $\begingroup$ I was kind of wondering that. Thanks for your answer. $\endgroup$ – muimerp Feb 17 at 19:52
  • $\begingroup$ A typo : 3rd last line g(c). $\endgroup$ – Peter Szilas Feb 17 at 19:53
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The last step follows from the transitive property of implication. From the second step you have $${|x-c|} < \delta \Rightarrow {|g(x)-g(c)|} < \delta_1$$ and from the first step you have $${|t-g(c)|} < \delta_1 \Rightarrow {|f(t)-f(g(c))|} < \epsilon$$ If we let $t=g(x)$, then we get the following chain of implications: $${|x-c|} < \delta \Rightarrow {|g(x)-g(c)|} < \delta_1 \Rightarrow {|f(g(x))-f(g(c))|} < \epsilon$$ This gives us that $f \circ g$ is continuous at $c$.

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  • $\begingroup$ Thanks. I should have thought about it. $\endgroup$ – muimerp Feb 17 at 20:34

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