If $L^*$ or $L^+$ is empty, can L be an infinite language?

Question

I have to prove or disprove the implications in these two situations

1. $L^* = \emptyset$ $\rightarrow$ $L$ is infinite
2. $L^+ = \emptyset$ $\rightarrow$ $L$ is infinite

Here are my thoughts.

1. 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?

2. 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?

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

1. $L$ is finite $\rightarrow$ $L^* \neq \emptyset$
2. $L$ 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.

Answer 1

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 $L^{+} = \emptyset$.