What is the counterexample in modal system K for “⬜A ➡A”?

As I understand it from Modal Logic 2.1 - the systems M, B, S4 & S5, I should not be able to show "⬜A ➡A" in modal logic K. The following truth tree seems to confirm this, if I did it correctly:

There is no contradiction in w0 and I have no possibly operator allowing me to continue the tree into a new world. Because the accessibility relation is not reflexive in K, I cannot derive A in w0 to reach a contradiction that I would be able to in modal system M.

The truth tree remains open, implying that I cannot show "⬜A ➡A" in K. However, when I try to find a counterexample in K, I am stuck.

Let v be the valuation function. There are two possibilities:

1. v(A) = 1. If that is the case then "⬜A" is 1 and so is "A" which means the conditional is true. So that valuation does not lead to a contradiction.
2. v(A) = 0. If that is the case then "⬜A" is 0 and so is "A" which means the conditional is again true.

I don't see how I can construct a counterexample in K. That counterexample should also work as a counterexample in M where this can be derived. Perhaps being invalid does not mean I can always create a counterexample, but without the counterexample is it really invalid?

Reference

Kane B channel, Modal logic 2.1 - the systems M, B, S4 & S5 https://youtu.be/VRVX7B5Iw14

• A valuation function in modal logic doesn't simply assign 1 or 0 to A, but rather it assigns 1 or 0 relative to a world. In w0, for example, ⬜A could get 1 while A gets 0, as your tree shows possible. – Eliran Nov 17 '18 at 16:22
• The semantics is v(□A,w)=T iff for every world w′ in W such that wRw' we have : v(A,w′)=T. If R is not reflexive, this means that not w0Rw0. Thus, as per previous comment, the fact that A is false in w0 does not contradict the fact that □A is true in it, because from this we cannot infer that A is true in w0. – Mauro ALLEGRANZA Nov 17 '18 at 16:41
• @MauroALLEGRANZA R is not reflexive and we only have world w0, so R is empty. I assume what you are saying is □A is true, A is false (because of ~A) and so the conditional is false and that would be the counterexample. – Frank Hubeny Nov 17 '18 at 16:58