Consider the given information.

Here,

\(\displaystyle{S}={\left\lbrace{10}{n}−{1}:{n}\in{Z}\right\rbrace}\)

And,

\(\displaystyle{U}={\left\lbrace{5}{m}+{4}:{m}\in{Z}\right\rbrace}\)

And, the set T is the set of odd integers.

So, the set is defined as,

\(\displaystyle{T}={\left\lbrace{2}{k}+{11}:{k}\in{Z}\right\rbrace}\)

Now, find the first \(\displaystyle{T}\cap{U}\).

Let k is the arbitrary element of S.

Thus,

\(\displaystyle{k}={10}{n}−{1}{n}\in{Z}\)

= 10n − 1 + 2 − 2

= (10n − 2) + 1

= 2 (5n − 1) + 1

Here, (5n - 1) is belongs to integers.

Let (5n - 1) is equal p. Put the value in the expression.

Thus,

\(\displaystyle{k}={2}{p}+{1}\in{T}\)

Thus, the number k belongs to T.

\(\displaystyle{k}\in{T}\)

Now, define for S.

k = 10n − 1

= 10n − 1 + 5 − 5

= (10n − 5) + 4

= 5(2n − 1) + 4

Let (2n - 1) is equal m. Put the value in the expression.

Thus,

\(\displaystyle{k}={5}{m}+{4}\in{U}\)

Thus, the number k belongs to S.

\(\displaystyle{k}\in{U}\)

From the above both the result, the given condition proved.

\(\displaystyle{S}\subseteq{T}\cap{U}\).

Here,

\(\displaystyle{S}={\left\lbrace{10}{n}−{1}:{n}\in{Z}\right\rbrace}\)

And,

\(\displaystyle{U}={\left\lbrace{5}{m}+{4}:{m}\in{Z}\right\rbrace}\)

And, the set T is the set of odd integers.

So, the set is defined as,

\(\displaystyle{T}={\left\lbrace{2}{k}+{11}:{k}\in{Z}\right\rbrace}\)

Now, find the first \(\displaystyle{T}\cap{U}\).

Let k is the arbitrary element of S.

Thus,

\(\displaystyle{k}={10}{n}−{1}{n}\in{Z}\)

= 10n − 1 + 2 − 2

= (10n − 2) + 1

= 2 (5n − 1) + 1

Here, (5n - 1) is belongs to integers.

Let (5n - 1) is equal p. Put the value in the expression.

Thus,

\(\displaystyle{k}={2}{p}+{1}\in{T}\)

Thus, the number k belongs to T.

\(\displaystyle{k}\in{T}\)

Now, define for S.

k = 10n − 1

= 10n − 1 + 5 − 5

= (10n − 5) + 4

= 5(2n − 1) + 4

Let (2n - 1) is equal m. Put the value in the expression.

Thus,

\(\displaystyle{k}={5}{m}+{4}\in{U}\)

Thus, the number k belongs to S.

\(\displaystyle{k}\in{U}\)

From the above both the result, the given condition proved.

\(\displaystyle{S}\subseteq{T}\cap{U}\).