# Why is a subset of a undecidable language decidable?

I have problems with the understanding why a subset of a undecidable language is decidable. We've proved in the lecture that $$HALT_T_M=$${$$$$|M is a TM and M halts on input w} is undecidable. Can someone give an example of a subset from $$HALT_T_M$$ and explain why it is decidable? Is $$\Sigma^*$$ a valid subset of $$HALT_T_M$$?

• $\Sigma^\ast$ is a superset of $HALT_{TM}$, $\varnothing$ is a subset and decidable. – ttnick Sep 3 at 12:14
• It is the 'hard' instances of a language that make a language undecidable. If you're familiar with algorithm analysis, you know that the running time of bubble sort is $O(n^2)$; however, on a fully sorted input, the running time is $n$ steps. The runtime is somewhat analogous to decidability/undecidability. Any instance $<M, w>$ of $HALT_{TM}$ makes a decidable set consisting of one element (because every finite set with elements from $\Sigma^*$ is decidable). However, we are interested in finding an algorithm which would solve the problem for all inputs, not just some subset of inputs. – diplodoc Sep 3 at 12:32
• Now I get it. $\emptyset$ and $\epsilon$ are regular and therefore decidable. Thanks. – BreadCrust Sep 3 at 13:15
• @BreadCrust $\epsilon$ is a string, not a set. – David Richerby Sep 3 at 15:06