Math4202 Topology II (Lecture 6)

Some intuitions:

By definition for partition of unity, consider the sets $W_i,V_i$ defined as

$$W_i=f^{-1}_i((\frac{1}{2n},1])\subseteq f^{-1}_i([\frac{1}{2n},1])\subseteq V_i=f^{-1}_i((0,1])\subseteq \operatorname{supp}(f_i)\subseteq U_i$$ $$V_1\subseteq \overline{V_1}\subseteq U_1$$

Note that $V_i$ is open and $\overline {V_i}\subseteq U_i$.

And $\bigcup_{i=1}^n V_i=X$.

and $W_i$ is open and $\overline{W_i}\subseteq V_i$.

And $\bigcup_{i=1}^n W_i=X$.

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Step 1:

$\exists$ V_i$ope subsets$i=1,\dots,n$such that$\overline{V_i}\subseteq U_i$, and$\bigcup_{i=1}^n V_i=X$.

For $i=1$, consider $A_1=X-(U_2\cup U_3\cup \dots \cup U_n)$. Therefore $A_1$ is closed, and $A_1\cup U_1=X$.

So $A_1\subseteq U_1$.

Note that $A_1$ and $X-U_1$ are disjoint closed subsets of $X$.

Since $X$ is normal, we can separate disjoint closed subsets $A_1$ and $X-U_1$.

So we have $A_1\subset V_1\subseteq \overline{V_1}\subseteq U_1$ (by normal space proposition ).

For $i=2$, note that $V_1\cup\left( \bigcup_{i=2}^n U_i\right)=X$,

Take $A_2=X-\left(V_1\cup\left( \bigcup_{i=3}^n U_i\right)\right)$ (skipping $U_2$).

Then we have $V_2\subseteq \overline{V_2}\subseteq U_2$.

For $i=j$, we have

$$A_j=X-\left(\left(\bigcup_{i=1}^{j-1}V_i\right)\cup \left(\bigcup_{i=j+1}^n U_i\right)\right)$$

and $\bigcup_{i=1}^n V_i=X$.

Repeat the above construction for $\{V_i\}_{i=1}^n$.

Then we have $\{W_i\}_{i=1}^n$ open and $W_i\subseteq \overline{W_i}\subseteq V_i\subseteq \overline{V_i}\subseteq U_i$.

And $\bigcup_{i=1}^n W_i=X$.

Step 2:

Using Urysohn’s lemma . To construct the partition of unity $\phi_i$.

Since $W_1\subseteq \overline{W_1}\subseteq V_1\subseteq \overline{V_1}\subseteq U_1$,

Note that $\overline{W_1}$ and $X-V_1$ are two disjoint closed subsets of normal space $X$

Then we can have $f_1:X\to[0,1]$ such that $f_1(\overline{W_1})=\{0\}$ and $f_1(X-V_1)=\{1\}$.

Then we have the remaining list of function $f_2,\dots,f_n$.

Recall the definition for support of functions $\operatorname{supp}(f_i)=\overline{\{x\in X: f_i(x)>0\}}$. Since $f_i(x)=0$ for $x\in X-V_i$, we have $\operatorname{supp}(f_i)\subseteq \overline{V_i}$

Next we need to check $\sum_{i=1}^n f_i(x)=1$ for all $x\in X$.

Note that $\forall x\in X$, since $\bigcup_{i=1}^n W_i=X$, then there exists $i$ such that $x\in W_i$, thus $f_i(x)=1$.

And $\sum _{i=1}^n f_i(x)\geq 1$.

Then we do normalization for our value. Set $F(x)=\sum_{i=1}^n f_i(x)$.

Since $F(x)$ is sum of continuous functions, $F$ is continuous.

Then we define $\phi_i=f_i/F(x)$, since $F(x)\geq 1$, we are safe to divide by $F(x)$ and $\phi_i(x)$ is continuous.

And $\operatorname{supp}(\phi_i)=\operatorname{supp}(f_i)\subseteq \overline{V_i}\subseteq U_i$.

And $\sum_{i=1}^n \phi_i(x)=\frac{\sum_{i=1}^n f_i(x)}{F(x)}=\frac{F(x)}{F(x)}=1$ for all $x\in X$.