Let {X_n} be a sequence of real numbers that is bounded above by M and such that X_n-->x Prove that x<=M

To prove that x is less than or equal to M, we need to use the definition of convergence and the fact that the sequence {X_n} is bounded above by M.

First, let's recall the definition of convergence: a sequence {X_n} converges to a limit x if, for any positive number ε, there exists an index N such that for all n ≥ N, |X_n - x| < ε.

Since {X_n} converges to x, we can choose ε = 1. By the definition of convergence, there exists an index N such that for all n ≥ N, |X_n - x| < 1.

Now, let's use the fact that {X_n} is bounded above by M. This means that for every term X_n in the sequence, X_n ≤ M.

Combining the two facts, for all n ≥ N, we have:

X_n ≤ M (1)
|X_n - x| < 1 (2)

To prove that x ≤ M, we can consider two cases:

Case 1: x < M.
Since x < M and X_n ≤ M for all n, we have X_n - x ≤ M - x. From equation (2), |X_n - x| < 1. Taking the absolute value, we have -(X_n - x) < 1. Rearranging, we get x - X_n < 1. Combining this inequality with X_n ≤ M (equation 1), we have x < M for all n ≥ N, which implies x ≤ M.

Case 2: x = M.
If x = M, then for every ε > 0, there exists an N such that for all n ≥ N, |X_n - x| < ε. Choosing ε = 1/2, we have |X_n - M| < 1/2. From X_n ≤ M (since the sequence is bounded above by M), we have -X_n ≤ -M. Combining this inequality with |X_n - M| < 1/2, we get -X_n < 1/2 - M. Rearranging, we have X_n > M - 1/2. Since X_n is bounded above by M, this is a contradiction.

Therefore, we conclude that x ≤ M.

Thus, we have proved that if {X_n} is a sequence of real numbers bounded above by M and X_n converges to x, then x ≤ M.