Could Someone please check my work

Let S be a nonempty bounded subset of IR and let I: 6 1R . Define k3 = {k3:3 E S}. Prove the following: W an. H..-\" .. 2 lfk > 0, then inf(kS) = k inf(S) wrs 11mg) 2 k - 1111(5) . . Suppose S is a nonempty bounded subset of R and k > 0. Since S is a nonempty bounded subset of R then inf(S) exists as a real number by the completeness axiom. Likewise, since k- S is a nonempty bounded subset of R then inf(k- S)exists as a real number by the ' completeness axiom. By definition 3.3.5, for the infimum, inf(S) is the greatest lower bound 1 of S, which means that for all s E S, s > inf(S). Since k > O, the previous inequality can be multiplied by k so that, k s > k- inf(S), which means k inf(S) is a lower bound of k3 by definition 3.3.2. Also by definition 3.3.5 for the infimum, inf(k- S) is the , greatest lower bound of k S, which means that for all ks 6 k5, ks _>_ inf(kS). Since 11: inf(S) is a lower bound of [ k 1111(5). ; Now, WTs inf(kS) s k 1111(5). i j Since inf(S) is the greatest lower bound of S, the number inf(S) + s for 5 > 0 is NOT a lower bound of S. Therefore, there exists a number, s' E S such that s' 0, the previous inequality can be multiplied by I: so that, k s inf(kS)ands' E S, thenk- s' > inf(kS) Combining the inequalities, k- 8' > inkS) and k s'