Set Theory Regarding Subsets (Easy Question) So this should be an easy question but I'm confused on the last 2 answer choices, shouldn't the last 2 choices both be subsets? However only one answer can be correct.
What answer choice is a subset of $\{1,2,3,4,5\}$
$1. \{2,4,6\}$
$2. \{1,2,3,4,...\}$
$3$. {$x$ exists in domain of Real Numbers $| 0 < x < 6$ AND $x$ is even}


*{$x$ exists in domain of Real Numbers $| 1 < x < 5$}


Obviously the first 2 choices are false but how in the world is either 3 or 4 not in considered a subset?
 A: Recall that $A$ is a subset of $B$ if every element of $A$ is an element of $B$.
So the answer should be a set whose elements are amongst $1,2,3,4,5$. In fact, since there are only finitely many elements in $\{1,2,3,4,5\}$, such subset must be a finite set itself.
So the options $2$ and $4$ are immediately off the table. What can you tell about the remaining two?
A: $$\{x\text{ exists in domain of Real Numbers }| 0 < x < 6\text{ AND $x$ is even}\}\\
=\{\text{set of all even }\mathbf{real\text{ }numbers}\text{ between }0,6\}$$
We see thus that
$$\{\text{set of all even }\mathbf{real\text{ }numbers}\text{ between }0,6\}\subsetneq\{1,2,3,4,5\}.$$I believe what you saw was that $$\{\text{set of all even }\mathbf{real\text{ }numbers}\text{ between }0,6\}\cap\{1,2,3,4,5\}\neq\emptyset$$
And
$$\{x\text{ exists in domain of Real Numbers }| 1 < x < 5\}\\
=\{\text{set of all }\mathbf{real\text{ }numbers}\text{ between }1,5\}$$
We see thus that
$$\{\text{set of all }\mathbf{real\text{ }numbers}\text{ between }1,5\}\subsetneq\{1,2,3,4,5\}.$$I believe what you saw was that $$\{\text{set of all }\mathbf{real\text{ }numbers}\text{ between }1,5\}\cap\{1,2,3,4,5\}\neq\emptyset$$
A: This can be answered very easily. First of all,  as you have understood first two options are false, let's go to next options.
4th option says x belongs to real numbers. But please note that if that's real numbers, then we can also include 1.1,1.2,1.3,......4.2,4.8,4.9,5.
But the question set doesn't mean it.
So, the correct answer is option 'C'
