We are asked to show that if $A$, $B$, and $C$ are subsets of the real numbers $\mathbb{R}$, and $f: A \to B$ and $g: B \to C$ are both continuous functions, then the composition $g \circ f : A \to C$ is also continuous.

Proof:

Let $f: A \to B$ and $g: B \to C$ be continuous functions. We need to show that the composition $g \circ f : A \to C$ is continuous.

By the definition of continuity, a function $h: X \to Y$ is continuous if for every open set $V \subset Y$, the preimage $h^{-1}(V)$ is open in $X$.

Step 1: Continuity of $g$ and $f$

• Since $f: A \to B$ is continuous, for any open set $U \subset B$, the preimage $f^{-1}(U)$ is open in $A$.
• Similarly, since $g: B \to C$ is continuous, for any open set $V \subset C$, the preimage $g^{-1}(V)$ is open in $B$.

Step 2: Composition $g \circ f$

Now consider the composition $g \circ f : A \to C$. For an open set $V \subset C$, we need to show that the preimage $(g \circ f)^{-1}(V)$ is open in $A$.

• By the definition of composition, we have: $(g \circ f)^{-1}(V) = f^{-1}(g^{-1}(V)).$ This means that $(g \circ f)^{-1}(V)$ is the preimage under $f$ of the set $g^{-1}(V)$.

Step 3: Open Set Preimage

Since $g$ is continuous, $g^{-1}(V)$ is open in $B$. And since $f$ is continuous, the preimage $f^{-1}(g^{-1}(V))$ is open in $A$.

Thus, $(g \circ f)^{-1}(V)$ is open in $A$ for any open set $V \subset C$. This shows that $g \circ f$ is continuous.

Conclusion:

Since $(g \circ f)^{-1}(V)$ is open in $A$ for any open set $V \subset C$, we conclude that the composition $g \circ f$ is continuous.

Would you like further details on any of the steps?

Here are 5 related questions you might explore:

1. What is the formal definition of a continuous function?
2. How does the concept of continuity generalize to metric spaces?
3. Can you provide examples of two specific continuous functions $f$ and $g$, and show that $g \circ f$ is continuous explicitly?
4. How do open sets in $\mathbb{R}$ relate to continuity?
5. Is the composition of uniformly continuous functions also uniformly continuous?

Tip: When proving continuity, it is often easiest to work with the preimage of open sets, as done here, due to its close connection with the definition of continuity.