# Problem understanding a step of a short proof

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?

• Has your question been answered? If yes, you should accept an answer. – Haris Gušić Jun 24 at 15:08

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 $$t = g(x)$$.
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 $$c$$.