If $L^*$ or $L^+$ is empty, can L be an infinite language? I have to prove or disprove the implications in these two situations


$L^* = \emptyset$ $\rightarrow$ $L$ is infinite
$L^+ = \emptyset$ $\rightarrow$ $L$ is infinite


Here are my thoughts.


I would say Kleene star operation (*) is union of 

$\qquad L^0 \cup L^1 \cup L^2 \cup \dots$

and since $\_^0 = \{\epsilon\}$ (even $\emptyset^ 0 = \{\epsilon\}$), there is no language $L$ such that $L^* = \emptyset$. 

How to prove finiteness/infiniteness?
Only language $L$ where $L^+ = \emptyset$ is $\emptyset$ which is a finite language. But how to prove there is no other language that is inifinite and the argument is valid for it?




EDIT: will it help if I rewrite the implications $A \rightarrow B$ to $\neg B \rightarrow \neg A$ ?


$L$ is finite $\rightarrow$ $L^* \neq \emptyset$
$L$ is finite $\rightarrow$ $L^+ \neq \emptyset$


2) can be now interpreted as $\emptyset^+ = \{\epsilon\}$ and $\emptyset$ is a finite language. So the implication doesn't work.
1) You need to take a look at this page. Remember that there is an implication to be proved true or false.

2) Consider the fact $L \subseteq L^{+}$ and apply it to the case $L^{+} = \emptyset$.