Answer
See explanation
Work Step by Step
**Answer assuming “bit strings” means all finite sequences of 0’s and 1’s:**
A standard argument is to observe that bit strings can be grouped by their length:
- There is exactly \(1\) string of length \(0\) (the empty string).
- There are \(2\) strings of length \(1\).
- There are \(2^2 = 4\) strings of length \(2\).
- \(\dots\)
- In general, there are \(2^n\) strings of length \(n\).
Thus, the set of **all** finite bit strings is the **countable union** of these finite sets:
\[
\underbrace{\{ \varepsilon \}}_{\text{length }0}
\;\cup\;
\underbrace{\{0,1\}}_{\text{length }1}
\;\cup\;
\underbrace{\{00,01,10,11\}}_{\text{length }2}
\;\cup\;\dots
\]
A countable union of finite sets is itself **countable**. Concretely, you can list all bit strings by writing first the length‐0 string, then all length‐1 strings, then all length‐2 strings, etc.:
\[
\varepsilon,\;
0,\;1,\;
00,\;01,\;10,\;11,\;
000,\;001,\;010,\;011,\;\dots
\]
Because we can arrange every finite bit string into a single infinite list (one after another), the set of all such strings is countable.
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**Note**: If instead one considers **infinite** bit strings (sequences of 0’s and 1’s of infinite length), that set is **uncountable**, akin to the cardinality of the real numbers in \([0,1]\) when viewed in binary form.
