Proof: is the language $L_1$$=${$|$$\emptyset \subseteq L(M)$} (un)-decidable?

I want to show that $ L_1$ $ =$ {$ <M>|$ $ \emptyset \subseteq L(M)$ } is decidable/undecidable – without rice theorem (just for the case that I can apply it).

Every language contain the $ \emptyset$ as a subset. So my guess is that the language is decidable.

Therefore, let us assume that $ L_1$ is decidable. Lets say that $ N$ is the TM which decides $ L_1$ .

N = "with input $ <M>$ :"

How can I proof that $ N$ is a decider for $ L_1$ ?

$L = \{ \langle M \rangle \mid M$ is a turing machine and $L(M)$ is decidable $\}$

$ L = \langle M \rangle \mid M$ is a turing machine and $ L(M)$ is decidable $ \}$ . How could I show this language is undecidable using the HALTing method?

A language L is undecidable if its not corecognizable and recognizable.

So I wish to show $ $ HALT \leq_m \overline{L}$ $ for not recognizable and

$ $ HALT \leq_m L$ $

Not corecognizable

SHOWING Not recognizable attempt:

R = "On input <M, w>     Construct M1     M1 = "On input x          run M on w          accept     return <M1> 

If $ \langle M, w \rangle \in HALT$ , then M1 accepts all input $ L(M_1) = \Sigma^*$ which is undecidable $ \langle M_1 \rangle \notin L$ as we would want

If $ \langle M, w \rangle \notin HALT$ , then M1 loops on all input $ L(M_1) = \emptyset$ , which is decidable $ \langle M_1 \rangle \in L$ as we would want

After further look into this $ L(M_1) = \Sigma^*$ is actually decidable? How would I do this then. What can I do to M1 to make it return an undecidable TM if M halts on w?

prove $ L=\{M\ |\ M\ is\ a\ TM\ and\ \forall.x\in \Sigma^*\ with\ |x|>2,\ M\ on\ x\ runs\ at\ most\ 4|x|^2\ steps\}\notin R$

I am trying to prove that the language

$ $ L=\{M\ |\ M\ is\ a\ TM\ and\ \forall.x\in \Sigma^*\ with\ |x|>2,\ M\ on\ x\ runs\ at\ most\ 4|x|^2\ steps\} $ $

belongs to $ Co-RE$ but not to $ R$ .

Showing $ \bar{L}\in RE$ is pretty much straight forward, but I also want to show that $ L\notin{R}$

My idea was a reduction $ \bar{H_{TM}}\le_mL$ but I struggle to figure out how to do it.

Any help/guidance will be much appreciated.

L: $\{\langle M \rangle \mid M$ is a TM and $\{111\} \subseteq L(M)\}$ basically 111 can get accepted by $L(M)$

L: $ \{\langle M \rangle \mid M$ is a TM and $ \{111\} \subseteq L(M)\}$ basically 111 can get accepted by $ L(M)$

Proof that this is recognizable:

Here is a recognizer $ M1$ for L:

Let M1 = "On input ⟨M⟩: 1. Run M on 111 2. if M accepts then accept else loop" 

Is this right? if $ \langle M \rangle \in L$ then it will accept $ \langle M \rangle \notin L$ then it will loop. I’m attempting to do this without dove tailing but cant find a lot of resources for this