# Classnotes for October 13

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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

## Basic Properties of Connected Sets

We recall, from last class, the definition of a connected set and a simple example:

Definition: A set is connected if it is not a non-trivial disjoint union of open sets.

Theorem: [0,1] is connected in $\mathbb{R}$.

As one might expect, continuous maps preserve connectedness.

Theorem: If $f\colon X \rightarrow Y$ is continuous, then X is connected implies that f(X) is connected.

Proof: If f(X) = $A \cup B$ with A and B disjoint and open, $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$ with f − 1(A) and f − 1(B) open and disjoint. That is, X is not connected with is a contradiction.

Intermediate Value Theorem: If $f\colon [a,b] \rightarrow \mathbb{R}$ where f(a) < 0 < f(b), then $\exists \; x \in [a,b]$ such that f(x) = 0.

Proof: f([a,b]) is connected since [a,b] is connected.

## Scanned Notes

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Handwritten notes by Annat Koren
Handwritten notes by Annat Koren
Handwritten notes by Annat Koren
Handwritten notes by Annat Koren