The intuition is that finite languages are very simple in the sense that throwing a finite number of words from a language does not affect membership in $text{R}$ or in $overline{text{RE}}$, etc. Therefore, since $L^{geq k}$ is the result of throwing a finite number of words from $L$, we get that:

- $L(M)^{geq k} in text{RE}$, and
- $overline{HP}^{geq k} notin text{RE}$.

So the language $L = { langle M rangle: exists k: L(M)^{geq k} = overline{HP}^{geq k}}$ must be empty.

Formally speaking, we have the following claims:

**Claim 1:** For every non-trivial language $A$, and every finite strict subset $Bsubsetneq A$, it holds that $A leq_m A setminus B$.

**Hint for the proof of claim 1:** $B$ is finite and thus decidable. Therefore, given input $x$ for the reduction, we can check whether $xin B$. Then you can proceed easily. Think where to map inputs $x$ from $B$, and where to map inputs $x$ from $overline{B}$.

We also have the following similar claim which could be useful.

**Claim 2:** For every non-trivial language $A$, and every finite strict subset $Bsubsetneq A$, it holds that $Asetminus B leq_m A $.

Given the above claims, we’re done. Indeed, $overline{HP} notin text{RE}$, and for every $k$, $overline{HP}^{geq k} = overline{HP}setminus {win overline{HP} : |w| < k}$. That is, $overline{HP}^{geq k}$ equals a non-trivial infinite language minus a finite subset, and so by claim 1, $overline{HP}^{geq k} notin text{RE}$. Also, it holds that $L(M)^{geq k} in RE$ for every machine $M$ (this is easy, you can prove that directly using standard closure properties and the fact that $L(M)^{geq k} = L(M)setminus { win L(M): |w| < k}$. Alternatively, you can use claim 2 but you have to be careful regarding the edge cases where $L(M)$ is trivial, etc.). Therefore, it cannot be the case that there is a machine $M$ with $L(M)^{geq k} = overline{HP}^{geq k}$.