Number of sets of minimal cardinality in a finite subcover of an open cover I am learning topological entropy on my own from a series of online notes, and the definition of entropy of an open cover $\alpha$ of $X$  is the following. The entropy of $\alpha$ is $H(\alpha)= log(N(\alpha))$, where $N(\alpha)$ is the number of sets in a finite subcover of $\alpha$ of minimal cardinality. (The space X is compact, hence N is always finite). 
Also for two open covers, we write $\alpha \leq \beta$ if for every $B\in \beta$ there exists an $A\in \alpha$ s.t. $B\subset A$. Also $\alpha V \beta= \{A\cap B: A \in \alpha, B \in \beta \}$
The following propositions are given without any proof. Here $\alpha$ and $\beta$ are two open covers of X. 
1) $H(\alpha)\geq 0$
2) $H(\alpha)= 0 \ $iff$ N(\alpha)=1 \ $iff$  X \in \alpha$
3) If $\alpha \leq \beta$, then $H(\alpha)\leq H(\beta)$
4)$ H(\alpha V \beta) \leq H(\alpha) + H(\beta) $
5) For $T: X->X$ continuous, $H(T^{-1}\alpha) \leq H(\alpha)$
I have 1,2 but need some help on the rest if that's possible. What is confusing me is how sub-covers relate to one another if we know how the covers relate to one another. For instance if for every set B in $\beta$ there exists an $A \in \alpha$, will that property carry on for the sets that end up in the subcovers? I am not convinced. Thanks in advance!
 A: The trick to showing that $H(\alpha)\leqslant H(\beta)$ for open covers $\alpha$ and $\beta$ is to show that, if we can find a finite subcover of $\beta$ of size $n$, then we can obtain a finite subcover of $\alpha$ of size $\leqslant n$.
Let's apply this to $(3)$. We have open covers $\alpha$ and $\beta$ with $\alpha\leqslant\beta$. Suppose $\{B_{1},...,B_{n}\}$ is a finite subcover of $\beta$ of size $n$, so $X=\cup_{i=1}^{n}B_{i}$. Since $\alpha\leqslant\beta$, for each $i$ we can find $A_{i}\in\alpha$ such that $B_{i}\subseteq A_{i}$. Thus, $X=\cup_{i=1}^{n}A_{i}$, so $\{A_{1},...,A_{n}\}$ is a finite subcover of $\alpha$ of size $n$. Thus, $N(\alpha)\leqslant n$ (since $N(\alpha)$ is defined to have minimal cardinality). Taking $n=N(\beta)$ shows that $N(\alpha)\leqslant N(\beta)$. Taking logs gives $H(\alpha)\leqslant H(\beta)$, as required.
A similar method can be used to prove the rest. For $(5)$, observe that $T(X)\subseteq X$. So if $\{A_{1},...,A_{n} \}$ is a finite subcover of $\alpha$ of size $n$, then for each $x\in X$, we can find $j$ such that $T(x)\in A_{j}$. Thus, as $T$ is continuous, this shows that $\{T^{-1}(A_{1}),...,T^{-1}(A_{n})\}$ is a cover of $X$. In particular, it is a finite subcover of $T^{-1}(\alpha)$ of size $n$. Thus, $N(T^{-1}(\alpha))\leqslant N(\alpha)$.
For $(4)$, a slightly different method is needed. Suppose $\{A_{1},...,A_{m} \}$ and $\{B_{1},...,B_{n} \}$ are finite covers of $\alpha$ and $\beta$ respectively. Then, $$X=\bigcup_{i=1}^{m}A_{i}=\bigcup_{j=1}^{n}B_{j}.$$
Hence, for any fixed $k\in\{1,...,m\}$, we deduce that $$A_{k}=A_{k}\cap X=\bigcup_{j=1}^{n}(A_{k}\cap B_{j}).$$ Taking the union over all $k$, we deduce that $$X=\bigcup_{k=1}^{m}A_{k}=\bigcup_{k=1}^{m}\bigcup_{j=1}^{n}(A_{k}\cap B_{j}).$$ Thus, $\{A_{i}\cap B_{j}:1\leqslant i\leqslant m,\, 1\leqslant j\leqslant n\}$ is a finite subcover of $\alpha \vee \beta$ of size $mn$. Hence, $N(\alpha \vee \beta)\leqslant mn$. Taking $m=N(\alpha)$ and $n=N(\beta)$ gives $N(\alpha \vee \beta)\leqslant N(\alpha)N(\beta)$. Taking logs gives $(4)$ (using the fact that $\log(xy)=\log(x)+\log(y)$).
I'm not sure if your notes mention this, but the $\vee$ notation is used to denote the fact that $\alpha \vee \beta$ is the join (i.e. least upper bound) of $\alpha$ and $\beta$. That is, $\alpha \leqslant \alpha\vee\beta$ and $\beta \leqslant \alpha\vee\beta$, and for any open cover $\gamma$ satisfying $\alpha\leqslant\gamma$ and $\beta\leqslant\gamma$, we have $\alpha\vee\beta\leqslant\gamma$.
