Closed or open set? And Cantor metric

Since this semester I started topology. I find this extremely difficult, it takes me much time to understand anything from it.

1. Please correct me if I am wrong. On $R^2$ space every continuous function is closed, because I am not able to insert a open ball into it with a positive radius.
2. Suppose we have a rectangle with $x \in [-2, 2]$ and $y \in (-\infty, \infty)$. I think this is a closed set, because its complement is open - again, because I can put open balls into the complement. Am I thinking right?
3. Our teacher told us about a metric on a Cantor set. Let $b$ and $c$ be sequences with elements either 0 or 1.

$d(b,c)=1/n$,
where n is the sequences index, at which $b$ and $c$ are different. If $b$ and $c$ are the same then $d(b,c)=0$.

Now I need to proof that this is a metric space. It's obvious that $d(b,c) \geq 0$ and $d(b,c)=d(c,b)$. But how to proof the triangle inequality? Three sequences can be different on different indexes.
• What do you mean by "insert an open ball into a continuous function"? Nov 24 '13 at 12:06
• You are talking about continuous functions on $\mathbb{R}^{2}$. That is a bit vague. Do you mean functions from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$ that are continuous when domain and codomain are both equipped with the usual producttopology? (i.e. the topology generated by sets like $\left(a,b\right)\times\left(c,d\right)$). Nov 24 '13 at 12:08
• You almost are right on point 2). The union of open balls that you can put into the complement must also coincide with the complement. Nov 24 '13 at 12:13
• I think he means the graph of a function $\mathbb R \to \mathbb R$. Nov 24 '13 at 12:19
• Yes, I mean a graph like $f(x) = sin x$. Nov 24 '13 at 18:21

1. The graph (http://en.wikipedia.org/wiki/Graph_of_a_function) of a continuous function is closed, but you've just shown that it's not open. Open and closed aren't opposites. A set can be both open and closed (like $\mathbb{R}^2$) or neither (like $(0,1]$). Try considering its complement (as you do for (2)). There's also a definition of closed based around the limits of sequences that would make things neater if you've covered it. And if you do it that way then as an extention you could think about what happens if you get rid of the requirement of continuity.
2. Assuming you mean an open ball around any point in the complement, then yes. As drhab says, just being able to fit an open ball into the complement isn't enough. For example $(0,1)$ isn't closed, but its complement contains $(3,4)$.
3. Let $a,b,c$ be sequences. If $a$ first differs from $b$ in the $m^{th}$ place and $b$ from $c$ in the $n^{th}$ place, what limits can you place on when $a$ first differs from $c$?

I believe that this site requires me to post actual answers rather than just hints, so spoilered answers are below.

(1)

Let $M$ be a metric space and $X$ a subset of $M.$ $X$ is closed if the limit of every sequence in $X$ that converges in $M$ is in $X.$ Let $(x_1,f(x_1)),(x_2,f(x_2)),\ldots$ be a sequence of points in the graph of $f.$ By continuity, if the $x_i$ converge to some $x \in \mathbb{R}$ then $f(x_i)$ converges to $f(x)$ and so the limit is in the graph as required. (You can do it from the "complement of an open set" definition, but it would be messier.)

(2)

Let $(x,y)$ be in the complement of your set. The open ball of radius $x-2$ (if $x$ is positive) or $-x-2$ (if $x$ is negative) with centre $(x,y)$ is contained entirely in the complement. Since we can do this for any point in the complement, the complement must be open and so the original set must be closed.

(3)

If $a$ first differs from $b$ in the $m^{th}$ place and $b$ from $c$ in the $n^{th}$ place, the $a$ first differs from $c$ in the first of these places at the earliest. Therefore $d(a,c) \leq max(1/n,1/m) = max(d(a,b),d(b,c)) \leq d(a,b) + d(b,c).$ So it is a metric space (and in fact an ultrametric space https://en.wikipedia.org/wiki/Ultrametric_space).

• Hints are perfectly acceptable; so is what you’ve done here. Nov 25 '13 at 3:48
• Brilliant. Thanks. Nov 25 '13 at 9:59