MATH 190A - Lecture 2

Review on

• Metric Space,
• Continuity,
• neighborhoods
• Open & Closed

Proposition (Continuity around Neighborhoods)

If $f$ is continuous at $a\in X⟺$ for every neighborhood $N$ of $f(a)$, the inverse image $f^{-1}(N)$ is a neighborhood of $a$.

Proof (Forward): Given neighborhood $N$ of $f(a)$, that given $B_{\epsilon }(f(a))$ for some $\epsilon >0$, then by continuity, $\exists \delta >0$ such that

$$B_{\delta }(a)\subseteq f^{-1}(B_{\epsilon }(f(a)))$$

which means

$$B_{\delta }(a)\subseteq f^{-1}(N)$$

so $f^{-1}(N)$ is a neighborhood of $a\in X$.

Proof (Backward): Given $B_{\epsilon }(f(a))$, this is a neighborhood of $f(a)$. Therefore by assumption,

$$f^{-1}(B_{\epsilon }(f(a)))$$

is a neighborhood of $a$. Hence, $\exists \delta >0$ such that

$$B_{\delta }(a)\subseteq f^{-1}(B_{\epsilon }(f(a)))$$

and it is continuous.

Theorem

A function

$$f:(X,d_1)\to (Y,d_2)$$

is continuous everywhere $⟺$ for every open set $U$ in $Y$, $f^{-1}(U)$ is open in $Y$.

Proof: See Proposition (Topological Characterization of Continuity).

In class, we proved it like this:

Given open $U$ in $Y$, take $a\in f^{-1}(U)$. Since $U$ is open, then it contains some $B_{\epsilon }(f(a))$ with $\epsilon >0$. By continuity of $f$ at $a$, we have $\exists \delta >0$ such that by the Triangle Inequality,

$$\begin{array}{rl}d(y,a)& \le d(y,x)+d(x,a)\\ & \le \delta +r\\ & \le (\epsilon -r)+r\\ & \le \epsilon \end{array}$$

such that

$$B_{\delta }(a)\subseteq f^{-1}(B_{\epsilon }(f(a)))\subseteq f^{-1}(U)$$

indeed, $f^{-1}(U)$ is a neighborhood of $a$. As $a$ arbitrary, this is true for $a\in f^{-1}(U)$.

Conversely, given $a\in X$ and $B_{\epsilon >0}(f(a))$ in $Y$, WTS $\exists B_{\delta }(a)\subseteq f^{-1}(B_{\epsilon }(f(a)))$. We know $B_{\epsilon }(f(a))$ is an open set, so its preimage must also be open in $X$. Thus, it must contain $B_{\delta }(a)$.