This question is continued from a previous thread I started, but it had more than one question so I had to move the other question here.

For this example consider an injective map $f: A \to 2^A$ then for each element $a\in A$ $\exists$ $f(a)\subseteq A$.

But does this also mean $f(a)\subseteq 2^A$?

I ask this because when writing out a simple numerical example of such a set, say $A=\{{ 7,8\}}$ then $2^A=\{{\varnothing,\{{ 7\}},\{{ 8\}},\{{7,8\}}\}}$, however since each $f(a)$ is an element of $2^A$ then we must have $f(a)\subseteq 2^A$ since any set $X$ is always a subset of itself: $X\subseteq X$ right?

Thank you.

Best Regards.


It can, but it usually isn't.

We always have $\varnothing\in2^A$ and also $\varnothing\subset2^A$. So there exists an element of the power set $2^A$ which is also a subset of the power set $2^A$.

On the other hand, in your example, $\{7\}\in2^A$, but $\{7\}\not\subset2^A$.

  • $\begingroup$ Thanks for your answer. Can you please explain in simple english why $\{7\}\not\subset2^A$.? $\endgroup$ – BLAZE May 22 '15 at 9:35
  • 1
    $\begingroup$ $\{7\}$ is not a subset of $2^A$ because there exists an element of $\{7\}$ which is not an element of $2^A$. That element is, of course, $7$. $\endgroup$ – Chris Culter May 22 '15 at 9:36
  • $\begingroup$ perfect, that's exactly what I needed to hear, thank you $\endgroup$ – BLAZE May 22 '15 at 9:39

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