Are the following arguments valid for why $C(K_1) \geq C(K_2)$?
There are several outcomes for the payoff.
- If $S<K_1$ then the payoff of option is $0$
- If $K_1<S<K_2$ then the payoff is $S−K_1$ which is positive
- If $K_1<K_2<S$ the the payoff is $-(S−K_2)−(S−K_1)$ which is $K_2−K_1$. Now since K2>K1 then this is positive.
This is a Type B arbitrage and so $C(K_1) \geq C(K_2)$ must hold
Exercise 1 (Monotonicity of the price of a call) Two call options on a stock are identical apart from the strike price. The value of a call option with strike price $K$ is denoted by $C(K)$. Use an arbitrage argument to show that $$ K_{1}<K_{2} \Rightarrow C\left(K_{1}\right) \geq C\left(K_{2}\right) . $$
An arbitrage exists if there is a portfolio of options with initial value $V_0 \leqslant 0$ , that is with a nonpositive cost of construction, such that upon expiration at time $T$ we have
$$\tag{*}\mathbb{P}(V_T \geqslant 0) =1, \quad \mathbb{P}(V_T > 0) > 0. $$
If $C(K_1) < C(K_2)$, then a portfolio that leads to arbitrage is long one option with strike $K_1$ and short one option with strike $K_2$. The initial cost here is $C(K_1) - C(K_2) < 0$. Using the payoff scenarios that you correctly specified, show that the arbitrage condition (*) is met. This requires that the probability of the stock price exceeding $K_1$ be greater than zero -- which is true under any reasonable stochastic model for the price of a common stock.
Note also that $C(K_1) = C(K_2)$ also leads to an arbitrage as the initial cost of the long-short portfolio is zero and you can make money from nothing with non-zero probability. To preclude arbitrage we must see $C(K_1) > C(K_2)$.
All of this assumes that you can transact at those prices, i.e., there is no bid-ask spread.