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Initially, $$\bigcap\limits_{n\in\mathbb{N}}(0,\frac{1}{n})$$ appears to be non-empty because there seems to be a positive real number that is really close to $0$ in the set.

However, we can prove that it is indeed an empty set by using the Archimedean Property.

How can i correctly visualize this set as an empty set?

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    Tell me what is your positive real number that is "really close to $0$". I'll find one even closer. – 2017-01-24
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    I don't really know exactly what you mean by "visualize this set as an empty set?" – 2017-01-24
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    @NeedForHelp but if u find one that is even closer to 0, then it is still in the intersection – 2017-01-24
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    @5xum in other words, in your mind, how do u see it as an empty set without proving that it is indeed an empty set formally. – 2017-01-24
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    @yh05 It's not "still in the intersection" The intersection is over all $n$. You're taking the limits in the wrong order. – 2017-01-24
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    @yh05 I think "huh, ok $0$ isn't in the set, and any number larger than $0$ also isn't in the set, so I guess it's empty". So, in a way, I sketch the proof in my head. – 2017-01-24
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    @yh05 The point I'm trying to make is that there is no (real) number that is the closest to $0$. Eventually, as $n$ grows, the right endpoint of the interval $(0,1/n)$ becomes closer and closer to $0$. To such an extent that given any "positive real number that is really close to $0$", call it $x$, I'll choose $N$ sufficiently large so that $1/N$ will be even closer to $0$ than $x$. So your number $x$ won't be in $(0,1/N)$, hence won't be in the intersection. Since this will be true no matter what $x$ you give me, you see that no $x$ can be in the intersection. – 2017-01-24
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    @NeedForHelp thanks so much. i misread your first comment. – 2017-01-24
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    @5xum u think so quickly, or maybe im just slow. i'll prefer NeedForHelp 's logic. – 2017-01-24
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    @yh05 Practice, practice, practice. This is your maybe fifth, maybe fiftieth time you encountered a problem like this. My count is probably closer to 1000. – 2017-01-24
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    @5xum yep, this is my first course in real analysis. i have to practice more! – 2017-01-24
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    I'm trying to see it, but I see nothing at all. – 2017-01-24

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For dealing with limits and infinite intersection, see this post: Infinite intersection and limits

The sequence $\left(\frac{1}{n}\right)$ matches the conditions of the post above. So we have:

$$ \lim\limits_{n\rightarrow \infty} \bigcap\limits_{n \in \mathbb{N}} \left(0,\frac{1}{n}\right) = (0,0) = \emptyset$$