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I will post the full question for context.

$A=${$(x, y)∈\mathbb{R^2}:|x|≤2$ and $|y| > 1$}, $B=${$(x, y)∈\mathbb{R^2}:x≥0$} and $C=A-B$, where $A, B$ and $C$ are subsets of the plane.

The question asks for diagrams of each, but I will just post a quick MS Paint sketch of C:

enter image description here

The question then asks me to simply write down the closure, interior and boundary of C in set notation for the Euclidean metric in $\mathbb{R^2}$ and the discrete metric, also in $\mathbb{R^2}$.

So my question is what difference does the metric make to this question? Do I have to run something through each metric? I can't find anything in my book going into anything for specific then a generic $d$. I'm not really looking for a answer to the question I'm doing, but more an explanation of what the question wants from me. I would really appreciate any insight and sorry in advance if this is a really stupid question with an obvious answer I've somehow missed.

My first attempt:

$Cl_{(\mathbb{R^2}, d^{(2)})}(C)=${$(x, y)∈\mathbb{R^2}:-2\leq x\leq 0, y\leq -1, y\geq 1$}

$Int_{(\mathbb{R^2}, d^{(2)})}(C)=${$(x, y)∈\mathbb{R^2}:-21$}

$Bd_{(\mathbb{R^2}, d^{(2)})}(C)=${$(x, y)∈\mathbb{R^2}:-21$}$\cup${$(x, y)∈\mathbb{R^2}:x=0, y<-1$}$\cup${$(x, y)∈\mathbb{R^2}:x=-2, y>1$}$\cup${$(x, y)∈\mathbb{R^2}:x=-2, y<-1$}.

I'm starting to think that closure, interior and boundary are the same for both metrics. If the closure is the smallest $d$-closed subset that contains $C$, and the discrete boundary is both open and closed, then I figure that it is just $C$ itself. Then that logic follows through with the interval and boundary hopefully.

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When talking about topological properties you need to know that different topologies (or different metrics) give different consequences even if we are working on the same set.

Example, if you take the set of real numbers and {0} as a subset, {0} is not open on the Euclidean metric, but it is so in the discrete metrics (notice: in discrete metric every subset is both closed and open).

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    I think I may be confused about what a metric/metric space is. I thought because metrics are distance functions, it requires at least 2 points to get a distance unless then point is on itself. Is {$0$} closed on the euclidean metric, because the only way to get $0$ from the metric is for $x=y$, and so it is in a set on its own, which includes itself? Thanks for the response! – 2017-02-26
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    In the metric spaces sure we measure the distance between two points, but when it comes to topological problems, as this is, we need to see them as topological space. On the metric space (X,d) we define A to be an open set if for any $x \in A$, there exists $\epsilon >0$ s.t. $\{ y \in X| d(x,y)<\epsilon \}$ is a subset of A. Then we define B to be closed in X if X\B is open. On your question, when we deal with the set {0}, we can take its complement i.e. R\{0} and we see that it is open on the Euclidean metric, so {0} is closed. – 2017-02-26
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    Thanks for clarifying that for me. I have had a shaky first attempt at the Euclidean metric closure, interval and boundary. I've edited it into my question, let me know if I'm on the right track if you have the time to spare. – 2017-02-26
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    Be careful on the boundary, there are some errors there! On the other hand, when you work on the discrete metric, every subset is clopen (closed and open), so IntC=ClC=C, but BdC=ClC\IntC=?. – 2017-02-26
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    Hopefully that's all the errors in the boundary for $d^{(2)}$. As for the boundary, in my book it says BdC=ClC-IntC is that the same as what you have put? I'm torn between Bd(c) being the empty set or the whole of C. I'm leaning towards the whole of $C$ though. – 2017-02-26
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    Yes, BdC=ClC-IntC. Now, knowing ClC and IntC in the discrete metric it is easy to find BdC :) – 2017-02-26
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    Bd(C)=C - C=0. Thank you, you've been amazingly helpful! – 2017-02-26
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let for a given distance "d" B(x,r) be open ball center x for r > 0

x is in the closure of C if $B(x,r) \cap C \ne \emptyset$ for all r > 0

x is in the interior of C if there exists r > 0 such that $B(x,r) \subset C$

x is in the frontier of C if $B(x,r) \cap C \ne \emptyset$ for all r > 0, and $B(x,r) \cap (\mathbb{R}^2 - C) \ne \emptyset$ for all r > 0

so I suggest you sketch what open balls look like for the metrics you need to consider, and then use above criteria to sketch the various regions

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    Didn't think of that, visualising it will help me a lot. Thank you! – 2017-02-26