Standard to Discrete

From 0506Topology

#	Week of...	Links (edit)
Fall
1	Sep 12	About, Tue, Thu, Std2Disc
2	Sep 19	Tue, Thu, HW1, 14 Sets
3	Sep 26	Tue, Thu, Photo
4	Oct 3	Tue, Thu
5	Oct 10	HW2, Tue, Thu
6	Oct 17	Tue, Thu, HW3
7	Oct 24	Mon, Tue, Thu
8	Oct 31	Tue, Thu, HW4
9	Nov 7	TE1, Tue, Thu
10	Nov 14	Tue, Thu, HW5
11	Nov 21	Tue, Thu
12	Nov 28	Tue, Thu, HW6
13	Dec 5	Tue, Thu
E	Dec 12	TE2
Spring
14	Jan 9	Tue, IT83, Thu, HW7
15	Jan 16	Tue, Thu
16	Jan 23	Tue, HW8, Thu
17	Jan 30	Tue, Thu
18	Feb 6	TE3, Tue, Thu
19	Feb 13	Tue, Thu
R	Feb 20
20	Feb 27	Tue, Thu, HW9
21	Mar 6	Tue, Thu, HW10
22	Mar 13	Tue, Thu
23	Mar 20	Tue, Thu, HW11
24	Mar 27	Tue, Thu
25	Apr 3	Tue, Thu, HW12
26	Apr 10	Tue, Thu
Study	Apr 17	Office Hours
Exams	Apr 24	Final, PM

We intend to classify all continuous functions from $(\mathbb{R}^n,\mathcal{T}_{std})$ to $(\mathbb{R}^n,\mathcal{T}_{dsc})$ . It will be demonstrated that this set of functions is equal to the set of constant-valued functions. First, however, some simple preliminaries:

Definition: Let $(X,\mathcal{T})$ be a topological space. Then a set $U \subset X$ is said to be closed (with respect to the topology $\mathcal{T}$ ) if $U c$ is open with respect to that topology.

Lemma: The only subsets of $\mathbb{R}^n$ that are both open and closed with respect to the standard topology are $\empty$ and $\mathbb{R}^n$ .

The proof of this lemma is found at the bottom of this page. With the above tools, we can now prove our main result:

Theorem: The set of continuous functions from $(\mathbb{R}^n,\mathcal{T}_{std})$ to $(\mathbb{R}^n,\mathcal{T}_{dsc})$ is the set of all constant-valued functions from $\mathbb{R}^n$ to $\mathbb{R}^n$ .

Proof: First we will show that any constant-valued function is continuous under these conditions. For any such function $f$ where $\forall x \in \mathbb{R}^n$ $f (x) = c 0$ , for any set $U$ , $f^{-1}(U) = \mathbb{R}^n$ if $c_0 \in U$ and $f^{-1}(U) = \empty$ otherwise. In either case, $f - 1 (U)$ is trivially open in $S$ and thus $f$ is continuous. (More generally, this proof can be adapted to show that a constant-valued function between any two topological spaces is always continuous.)

Now let $f\colon (\mathbb{R}^n,\mathcal{T}_{std}) \to (\mathbb{R}^n,\mathcal{T}_{dsc})$ be continuous. Since every set $U \subset \mathbb{R}^n$ is open in $\mathcal{T}_{dsc}$ , by the continuity of $f$ we have that $\forall U \subset \mathbb{R}^n \ f^{-1}(U) \in \mathcal{T}_{dsc}$ . Let $y \in \mathbb{R}^n$ be arbitrary. If $A = f - 1 (y)$ then $A$ is open in $\mathcal{T}_{std}$ . Also, $B = f^{-1}(\mathbb{R}^n \setminus \{y\})$ is open in $\mathcal{T}_{std}$ . However, $B = f - 1 ({y} C) = (f - 1 ({y})) C = A C$ . Since the complement of $A$ is open, $A$ is closed in $\mathcal{T}_{std}$ . By our lemma, $A$ must be either $\empty$ or $\mathbb{R}^n$ . Since either no points or all points are mapped to $y$ , for any $y$ , $f$ is a constant function, q.e.d.

Proof of Lemma: Suppose, for contradiction, that $U \subset \mathbb{R}^n$ is open and closed with respect to the standard topology, and is neither equal to $\empty$ nor $\mathbb{R}^n$ . Let $D =$ $\{|x-y| : x \in U , y \in U^c\}$ . $D$ is a non-empty set (since both $U$ and $U c$ are non-empty) of real numbers bounded from below by zero and thus has an infimum. Thus, let $a \in U$ and $b\in U^c$ be such that $| a - b |$ is minimal. Note that if we find another point $p \in U$ such that $| p - b | < | a - b |$ then we have found a contradiction.

Since $U$ is open and $a \in U$ , we can find an $α > 0$ such that $B_\alpha(a) \subset U$ . Consider the continuous function $g\colon [0,1] \to \mathbb{R}^n$ defined by $g (t) = (1 - t) a + t b$ . Note that for $t \in (0,1]$ , $g (t)$ is closer to $b$ than $a$ is: $| g (t) - b | =$ $| (1 - t) a + t b - b |$ $= | a - b | | 1 - t | < | a - b |$ .

Since $g$ is continuous, $\exists \delta > 0$ such that $g(B_\delta(a) \cap [0,1]) \subset B_\alpha(a)$ . Therefore $g( \begin{matrix} \frac{\delta}{2} \end{matrix}) \in B_\alpha(a) \subset U$ . This contradicts that $| a - b |$ is minimal, q.e.d.

1) Let us show that constant maps are continuous.

Take $f(x)=y_0 \ \forall x\in (R^n,std)$ , and let $B \subset (R^n,discr)$

$\Rightarrow f^{-1}(B)= R^n$ if $y_0 \in B$

$f^{-1}(B)=\empty$ else

Both $R n$ and $\empty$ are open sets in $(R n, s t d)$ , so a constant function is continuous.

2) Suppose $f$ to be continuous from $(R n, s t d)$ to $(R n, d i s c r)$ .

Take any point $y\in (R^n,discr)$ , and its inverse image $A = f^{-1}(y),\ \ A \subset (R^n,std)$ . $A$ must be open if we want $f$ to be continuous. Now consider $B\subset (R^n,discr),B=R^n-\{y\}$ , which is open because it is a union of open sets (the points), and its inverse image $f - 1 (B) = R n - A$ . $(R n, s t d) - A$ is closed, because the complementar sets of open sets are closed, and the only open sets in $(R n, s t d)$ which are both open and closed are $R n$ and $\empty$ , so if we want $f - 1 (B)$ to be open id est $f$ to be continuous we have to take either $A=\empty$ or $A = R n$ , and $f$ is a constant again.

If you think that's cheating because we never defined closed sets, we'll try to show that $R n - A$ can not be open in the standard topology if A is open, $A\neq \empty, A\neq R^n$ . Take any $x \in A$ (you can do it because A is non empty), and a point y which does not belong to A s.t. $d(x,y)=\inf_{z, z\neg\in A} d(x,z)$ .(A priori, it could be impossible to find such y which also does not belong to A, but A is open, so if you take a sequence of point whose distance tends to the infimum, they're bounded because theyre eventually inside a circle of radius infimum+ $ε$ around A, so they accumulate to a point which reaches the infimum, which can not belong to A because there is a sequence of points outside A which tends to it so you cannot find an open ball contained in A). We could take the infimum because the set $R n - A$ is non empty. Now consider the segment S between x and y. Except for y, $S\subset A$ , because all its points have distance from x less than y, which attends the infimum between the points which do not belong to A. Now we can find along the segment a sequence ${x_n}\in S$ , ${x_n}\rightarrow y$ : this implies $\neg\exists B_\epsilon (y) \subset R^n-A$ , because $\forall \ \epsilon \ \exists \ \ x_{n_\epsilon}\in A, d(x_{n_\epsilon},y)<\epsilon$ , so the complementar set of A can not be open unless he, or A, are empty.

By the way, does anybody know how to transform the double three plug into a thorus?

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Standard to Discrete

From 0506Topology

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