Weaker and Stronger Topologies

# Weaker and Stronger Topologies

Definition: Let $X$ be a topological space and let $\tau_1$ and $\tau_2$ be topologies on $X$. Then $\tau_1$ is said to be Weaker or Coarser than $\tau_2$ if $\tau_1 \subseteq \tau_2$, and $\tau_1$ is said to be Stronger or Finer than $\tau_2$ if $\tau_1 \supseteq \tau_2$. |

If $\tau_1$ is weaker than $\tau_2$ then openness, closedness, compactness, and continuity of a function on $f$ all with respect to $\tau_1$ implies openness, closedness, compactness, and continuity of $f$ all with respect to $\tau_2$. This is summarize in the proposition below.

Proposition 1: Let $X$ be a topological space and let $\tau_1$ and $\tau_2$ be topologies on $X$. Suppose that $\tau_1$ is weaker than $\tau_2$. Then:a) If $O$ is open with respect to the topology $\tau_1$ then $O$ is open with respect to the topology $\tau_2$.b) If $C$ is closed with respect to the topology $\tau_1$ then $C$ is closed with respect to the topology $\tau_2$.c) If $K$ is compact with respect to the topology $\tau_2$ then $K$ is compact with respect to the topology $\tau_1$.d) If $f : X \to \mathbb{C}$ is continuous with respect to the topology $\tau_1$ then $f$ is continuous with respect to the topology $\tau_2$.e) If $(x_n)$ converges to $x$ with respect to the topology $\tau_2$ then $(x_n)$ converges to $x$ with respect to the topology $\tau_1$. |

**Note that (a) and (b) says if $O$ ($C$) is open (closed) in a topology then $O$ ($C$) is open (closed) is any stronger topology. (d) says that if $f$ is continuous with respect to a topology then it is continuous with respect to any stronger topology. (c) is different. It says that if $K$ is compact with respect to a topology then it is compact with respect to any WEAKER topology!**

**Proof of a)**Suppose that $O$ is open with respect to the topology $\tau_1$. Then $O \in \tau_1$. Since $\tau_1 \subseteq \tau_2$ we have that $O \in \tau_2$. So $O$ is open with respect to the topology $\tau_2$. $\blacksquare$

**Proof of b)**Suppose that $C$ is closed with respect to the topology $\tau_1$. Then $C^C$ is open with respect to $\tau_1$. So $C^C \in \tau_1$. Since $\tau_1 \subseteq \tau_2$ we have that $C^C \in \tau_2$. So $C$ is closed with respect to the topology $\tau_2$. $\blacksquare$

**Proof of c)**Suppose that $K$ is compact with respect to the topology $\tau_2$. Let $\{ O_i : i \in I \}$ be an $\tau_1$-open cover of $K$. Since each $O_i \in \tau_1$ and $\tau_1 \subseteq \tau_2$ we have that each $O_i \in \tau_2$. So $\{ O_i : i \in I \}$ is also a $\tau_2$-open cover of $K$. Since $K$ is compact with respect to the topology $\tau_2$, $O$ has a finite subcover, say $U = \bigcup_{n=1}^{\infty} I_n$. Then $\{ O_i: 1 \leq i \leq n \}$ is a $\tau_1$-open subcover of $\{ O_i : i \in I \}$ and thus $K$ is compact with respect to the topology $\tau_1$.

**Proof of d):**Let $f : X \to \mathbb{C}$ and let $U \subseteq \mathbb{R}$ be open. Since $f$ is continuous with respect to the topology $\tau_1$ we have that $f^{-1}(U) \in \tau_1$. But then since $\tau_1$ is weaker than $\tau_2$ we have that

\begin{align} \quad f^{-1}(U) \in \tau_1 \subseteq \tau_2 \end{align}

- So $f$ is continuous with respect to the topology $\tau_2$. $\blacksquare$

**Proof of e)**Since $(x_n)$ converges to $x$ with respect to the topology $\tau_2$ we have that for all $U \in \tau_2$ with $x \in U$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $x_n \in U$. Since $\tau_1 \subseteq \tau_2$ this means that for all $U \in \tau_1$ with $x \in U$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $x_n \in U$. So $(x_n)$ converges to $x$ with respect to the topology $\tau_1$.