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Why is the following true?
$$\{ \emptyset \} \neq \emptyset $$
Is it just because the Empty set already refers to there being a set present and so it is just the same argument that $\{ \{3\} \} \neq \{ 3\} $?
And is there a proof for the following?
For any set$\, X$,
$$X \cup \emptyset = X \quad \text{and} \quad X \cap \emptyset = \emptyset$$
Two sets are identical if and only if they have exactly the same elements. $\{\emptyset\}\not = \emptyset$ because they don't have the same elements: The first set has a set as an element (namely the empty set), the second one has no elements at all - thus they can't be identical.
More formally, the cardinality (i.e. the number of elements) of $\{\emptyset\}$ is $1$ ($|\{\emptyset\}| = 1$), while the cardinality of $\emptyset$ is $0$ ($|\emptyset| = 0$). If two sets already differ in the number of elements they have, they can not be the same.
To determine the cardinality of a set, it isn't of interest what the elements of the respective set contain (i.e., that $\emptyset$ as an element of $\{\emptyset\}$ contains no further elements), but what the set itself contains - you only need to look at what is immediately inside the outer brackets.
In this case, if you re-write $\emptyset$ as $\{\}$ and $\{\emptyset\}$ as $\{\{\}\}$ and then look at the content of the outmost brackets, you will see that $\{\{\}\}$ contains the element $\{\}$, while $\{\}$ contains nothing at all.
As for your question,
Is it just because the Empty set already refers to there being a set present and so it is just the same argument that $\{ \{3\} \} \neq \{ 3\} $?
yes,that's exactly the reason. With $\emptyset$ being a content of $\{\emptyset\}$, there already is "a set present" (and thus an entity present in the outer set), and this is the same as the fact that a set containing a set (which further contains a number) is not the same as a set containing a number.
It might help you to imagine sets as boxes with stuff in it. So, $\{apple, orange, \{pen\}\}$ would be a box with an apple, an orange, and another smaller box in which there is a pen.
$\{\emptyset\}$ is a box with an empty box inside; when you open the outer box, you will be happy to see that it has something in it - namely a box, and that you might get disappointed about this inner box being empty is irrelevant to the fact that the outer box itself that we are talking about does have something in it, namely a box.
$\emptyset$, on the other hand, is really just an empty box: If you open it, you will find nothing. And an empty box is obviously not the same as a box in which there is a box.
As for the second part of your question:
The union between two sets $X$ and $Y$ is the set of all elements that are in either $X$ or $Y$. More formally, $X \cup Y = \{x : x \in X\text{ or }x \in Y\}$.
This means that you go and "collect" everything from both $X$ and $Y$ and put it together in one set. Now it is clear that this set will contain everything which is in $X$, but if the second set is the empty set, there will be no elements to further add to the union of the two sets. Thus, you will end up with only the contents of set $X$, since $\emptyset$ can not contribute any to it.
This means that unifying any set $X$ with the empty set $\emptyset$ will result in $X$ again; formally, $X \cup \emptyset = X$.
To pick up the box example again, you take a box $X$, pour all its content into a new box, and then additionally pour the content of an empty box into it. Unsurprisingly, your newly created union box will contain exactly what was in box $X$ before.
A similar reasoning applies to the intersection between two sets $X$ and $Y$, which is the set of all elements that are contained both in $X$ and in $Y$, formally $X \cap Y = \{x : x \in X\text{ and }x \in Y\}$. You can informally also word it as "the set of all elements that $X$ and $Y$ share".
What you are more or less doing here is taking out each element of set $X$ and check whether it is also contaned in $Y$ (or vice versa); if so, this element will be part of their intersection, otherwise not. Now which elements does any set $X$ share with the empty set? Not much: Since the latter has no elements at all, the search for any element which is in $\emptyset$ and also in $X$ is of no avail, you will end up with an empty set again.
So for any set $X$, the intersection with the empty set $\emptyset$ will be an empty set, as the two sets share no common elements (because $\emptyset$ has no elements at all); formally, $X \cap \emptyset = \emptyset$.
In terms of boxes, you create a new box in which you put all the things that you have twice, once in box $X$ and once in the empty box. You will be done very quickly: As the second box is empty, there is nothing to put into your box of shared elements, thus, your intersection box will remain empty.
Your intuition is correct. $\{\emptyset\}$ can be thought of as the set with one element, namely the empty set. Similarly for any set $A$, $\{A\}$ is the set containing just one element, the set $A$. You can think of $\emptyset$ just as something containing nothing, an empty container to be dangerously colloquial.
I'm assuming you are not familiar with axiomatic set theory judging from your tags, so I'll present a proof using naive set theory for your first question, and you can follow my style of proof for the second.
If $X$ is empty the result is vacuously true, so assume it isn't. Choose an arbitrary $x\in X$, then $x$ is contained in $X$ so it is contained in $X$ or $\emptyset$. Hence $x\in X\cup\emptyset$. As $x$ was arbitrarily chosen $X\subseteq X\cup\emptyset$. Now choose an arbitrary $y\in X\cup\emptyset$. Then $y$ is contained in $X$ or $y$ is contained in $\emptyset$. Obviously the latter is impossible, because nothing is in an empty container, so $y\in X$. As above it follows that $X\cup\emptyset\subseteq X$. So by the both-ways containment the desired equality must follow.
Here is some intuition for you: The simplest explanation is that the set containing the empty set is obviously not empty because it has the empty set inside of it. The literal mathematical meaning of this is {$\varnothing$}$\neq\varnothing$.
For the second one? What is the meaning for the union of two sets, other than the combination of both of them together into one set? Think: If you combine a set with a set that has nothing it, then you get back the same set as the first. That that lousy empty one didn't contribute anything at all.
What about the 3rd one? What's the meaning of the intersection of two sets other than the collection into a new set of all the things they held in common? Now think: Given any set and an empty set, mentally collect all things present in both. To be in the new collection, they have to have been present simultaneously in both. Did you get any objects that were present in both sets? No. Why? You got ripped off. One set had nothing in it there could not be any objects present simultaneously in both. You should probably go get your money back or something. Your good non-empty set vanished.
Okay, since you got ripped off on the last one, I'll give you something for free as a bonus:
Why do we call it "the" empty set, rather than simply an empty set? Because there is only one empty set. What's the reason?
Well, if you have two empty sets, it must be that they're actually identical and thus one because they all contain exactly the same thing: nothing! Here's why: if we are talking about two different people, and you suddenly realize that they are exacly the same in every way possible, even more identical than twins could be, then they must be the same person and you...you must have been tripping or something....all talking about him like he was two people. wow....
Awful weird isn't it?
Adam V. Nease
The axiom of extensionality says that two sets are equal if and only they have the same elements. More precisely, $A=B$ if and only if "for every $x$, $x\in A$ if and only if $x\in B$". (Quotes are used as delimiters for more complex statements.)
The statement $\emptyset\in\{\emptyset\}$ is true; on the other hand, $x\in\emptyset$ is false for every $x$, in particular for $x=\emptyset$.
Thus, yes: $\{\emptyset\}\ne\emptyset$.
For the second part, recall that, by definition, $A\subseteq B$ if and only if "for every $x$, if $x\in A$, then $x\in B$". Thus the axiom of extensionality can be rewritten as
$A=B$ if and only if "$A\subseteq B$ and $B\subseteq A$"
Now the definition of set union has the consequence that $A\subseteq A\cup B$. In particular, $X\subseteq X\cup\emptyset$, so we just have to show that $$ X\cup\emptyset\subseteq X $$ Let $x\in X\cup\emptyset$. So $x\in X$ or $x\in\emptyset$. Since the latter is false, only $x\in X$ remains.
Similarly, the definition of intersection tells us that $A\cap B\subseteq B$. In particular $X\cap\emptyset\subseteq\emptyset$. Since $\emptyset$ is a subset of every set, we have also $\emptyset\subseteq X\cap\emptyset$. Hence $X\cap\emptyset=\emptyset$.
You have several possible ways to denote a specific set. One of these is to write explicitly the elements contained in your set between braces. For example,
$$A=\{2,3,4 \}$$
is the set containing the natural numbers $2,3$ and $4$. Then, if the symbol $\emptyset$ denotes the set with no elements, it is clear that this is different from the set $\{ \emptyset \}$, which contains one element.
For what concerns your second question, you just have to think about the definitions of union and intersection of sets: if $X$ is any set, what are the elements contained in $X$ OR in $\emptyset$? And what about the elements contained in $X$ AND in $\emptyset$?
