"$\bigcap_{i\in I} A_i \subseteq A_k$, for all $k\in I$. If $B \subseteq A_k$, for all $k\in I$, then $B \subseteq \bigcap_{i\in I} A_i.$"

Proof. Let $k_{0}$ be an arbitrary element of $I$. Suppose $x\in B$, by hypothesis $x\in A_{k_0}$. Note that $k_0$ is an arbitrary element of $I$, then $x\in A_{k}$, for all $k\in I.$ Therefore $x\in \bigcap_{i\in I} A_i. \Box$

"$A_k \subseteq \bigcup_{i\in I} A_i$, for all $k\in I$. If $A_k \subseteq B$, for all $k\in I$, then $\bigcup_{i\in I} A_i \subseteq B$."

Proof. Suppose $x\in \bigcup_{i\in I} A_i$, then $x\in A_i$, for at least one $i\in I$. Let $k_0\in I$ such that $x\in A_{k_0}$. Given that $x\in A_{k_0}$, and $A_k \subseteq B$, for all $k\in I$, then $x\in B$. Hence $\bigcup_{i\in I} A_i \subseteq B. \Box$

Are my proofs correct? English is not my native language, if there's something wrong with my use of the language (inside the proof) I'd like to now it. Thank you!

  • $\begingroup$ Looks good to me. $\endgroup$
    – user370967
    Apr 14, 2017 at 8:42
  • $\begingroup$ Hello, @Math_QED. I have a question related to my second proof. Do you think replacing "Let $k_0\in I$ such that $x\in A_{k_0}$" with "Let $k_0\in I$, thus $x\in x_{k_0}$" (taking for granted $k_0$ is an arbitrary element of $I$) would be acceptable? $\endgroup$
    – mathman
    Apr 14, 2017 at 9:21
  • $\begingroup$ This seems weird. I added an answer so you see how I would prove it. $\endgroup$
    – user370967
    Apr 14, 2017 at 10:09
  • $\begingroup$ Btw, good job that you wrote your first post in such high quality (good formatting etc.) +1 $\endgroup$
    – user370967
    Apr 14, 2017 at 10:21

1 Answer 1


Your proofs both are correct. What looks a little strange (but not entirely wrong), is that you choose $k_0 \in I$.

I will prove both statements now as I would do it. You will notice both proofs are nearly exactly the same as yours are.

1) Let $x\in B$. Then, we know that $x \in A_k $ for all $k \in I$. Therefore, $x \in \bigcap_{i\in I} A_i$ (if an element is in all sets $A_i$, then certainly in their intersection). We conclude that $B \subset \bigcap_{i\in I} A_i.\quad\triangle$

2) Let $x \in \bigcup_{i \in I}A_i$ Then, there exists a $l \in I$ such that $x \in A_l$ (if an element is in a union, then the element is certainly in one of the sets forming the union). Because $A_l \subset B$ (as this is true for all sets $A_k$, certainly true for the specific set $A_l$), we find that $x \in B$. Hence, $\bigcup_{i \in I}A_i \subset B.\quad\triangle$

The technique is always the same. If you want to show that $A \subset B$,take an element $a \in A$ and show $a \in B$.


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