answer:It's not possible to write a function that checks if a sentence is a pangram with a constant time complexity, because you must at least read the entire sentence once, which is a linear time operation. However, you can write a function that has a constant space complexity and a linear time complexity. Here is a Python function that does that: ```python def is_pangram(sentence): seen = [False] * 26 for c in sentence: if c.isalpha(): seen[ord(c.lower()) - ord('a')] = True return all(seen) ``` This function works by creating a boolean array `seen` where each index represents a letter of the alphabet. As it iterates over the sentence, it sets the corresponding index in `seen` to True if it encounters a letter. Finally, it checks if all indices in `seen` are True using the `all` function. If they are, the sentence is a pangram. Note that this function assumes that the input sentence is encoded in a character set where the letters 'a' through 'z' have consecutive code points, which is the case for ASCII and Unicode. You can test the function with the same examples: ```python print(is_pangram('The quick brown fox jumps over the lazy dog')) # True print(is_pangram('Hello, world!')) # False ```

answer:Here is a Python function that checks if a sentence is a pangram for any given alphabet: ```python def is_pangram(sentence, alphabet): seen = {c: False for c in alphabet} for c in sentence: if c in alphabet: seen[c] = True return all(seen.values()) ``` In this function, the `alphabet` parameter is a string that contains all the characters of the desired alphabet. The function creates a dictionary `seen` where each key is a character in the alphabet and each value is initially False. As it iterates over the sentence, it sets the value in `seen` to True if it encounters a character in the alphabet. Finally, it checks if all values in `seen` are True using the `all` function. If they are, the sentence is a pangram for the given alphabet. You can test the function with a custom alphabet like this: ```python print(is_pangram('The quick brown fox jumps over the lazy dog', 'abcdefghijklmnopqrstuvwxyz')) # True print(is_pangram('Hello, world!', 'abcdefghijklmnopqrstuvwxyz')) # False print(is_pangram('abcdefgh', 'abcdefgh')) # True print(is_pangram('abcd', 'abcdefgh')) # False ``` If you want the function to be case-insensitive, you can modify it to convert both the sentence and the alphabet to lowercase or uppercase: ```python def is_pangram(sentence, alphabet): alphabet = alphabet.lower() seen = {c: False for c in alphabet} for c in sentence.lower(): if c in alphabet: seen[c] = True return all(seen.values()) ``` This way, the function will treat 'A' and 'a' as the same character. Also, if you want to ignore non-alphabetic characters in the alphabet, you can modify the function to filter them out: ```python def is_pangram(sentence, alphabet): alphabet = ''.join(c for c in alphabet if c.isalpha()).lower() seen = {c: False for c in alphabet} for c in sentence.lower(): if c in alphabet: seen[c] = True return all(seen.values()) ```

answer:To show that the union of two non-overlapping open intervals is not an open interval, let's consider the intervals (a, b) and (c, d), where a < b, c < d, and b ≤ c. We claim that the union of these two intervals, (a, b) ∪ (c, d), is not an open interval. To see this, recall the definition of an open interval: a set of real numbers I is an open interval if there exist real numbers a and b, with a < b, such that I = {x ∈ ℝ | a < x < b}. By definition, an open interval must be connected, meaning that it cannot be written as the union of two non-empty, non-overlapping open sets. Now, since b ≤ c, there is no real number x that belongs to both (a, b) and (c, d). Hence, the union (a, b) ∪ (c, d) is not connected and therefore cannot be an open interval. However, this does not imply that the union of two disjoint open intervals is a counterexample to show that the union of any two open sets is not an open set. In fact, the union of two disjoint open intervals is still an open set, just not an open interval. In general, the union of any two open sets is an open set, as it can be shown using the definition of open sets in terms of neighborhoods. In a topological sense, an open set is a set U such that for each point x in U, there exists an open neighborhood V of x such that V ⊆ U. Given two open sets U and V, their union U ∪ V is also an open set because for any point x in U ∪ V, we can find an open neighborhood of x that is contained in either U or V, and hence in U ∪ V. In summary, the union of two non-overlapping open intervals is not an open interval, but it is still an open set. This highlights the distinction between the concepts of open intervals and open sets in real analysis.

answer:Consider the two disjoint open intervals (0, 1) and (2, 3). We want to show that their union, (0, 1) ∪ (2, 3), is an open set. To do this, we need to show that for any point x in (0, 1) ∪ (2, 3), there exists an open neighborhood of x that is contained in (0, 1) ∪ (2, 3). Let's consider two cases: x ∈ (0, 1) and x ∈ (2, 3). If x ∈ (0, 1), then we can choose an open neighborhood of x, say (x - ε, x + ε), such that ε is small enough so that (x - ε, x + ε) ⊆ (0, 1). For example, we can take ε = min{x, 1-x}. Then (x - ε, x + ε) is an open interval that contains x and is contained in (0, 1). Similarly, if x ∈ (2, 3), we can choose an open neighborhood of x, say (x - ε, x + ε), such that ε is small enough so that (x - ε, x + ε) ⊆ (2, 3). For example, we can take ε = min{x-2, 3-x}. Then (x - ε, x + ε) is an open interval that contains x and is contained in (2, 3). In either case, we have found an open neighborhood of x that is contained in (0, 1) ∪ (2, 3). Therefore, (0, 1) ∪ (2, 3) is an open set. This example illustrates that the union of two disjoint open intervals is indeed an open set, even though it is not an open interval.