1.9 KiB
lecture, date
| lecture | date |
|---|---|
| 8 | 2025-02-03 |
Defines in Topological Space:
Proposition
[0, 1] is connected.
Proof
[0, 1] = A \cup B
A \cap B = \emptyset
Suppose Open Sets A and B disconnected it, and say 1 \in B.
Then every neighbourhood of a \equiv \sup{A} \in [0, 1] will intersect both A and B.
Which is a contradiction as this is impossible!
QED.
2.2 Continuity
Recall: f: \mathbb{R} \to \mathbb{R} is continuous at x if \lim_{ y \to x }f(y) = f(x), and it would be discontinuous if \lim_{ y \to x } f(y) \neq f(x). More precisely, if \forall \varepsilon \gt 0 \, \exists \delta \gt 0 such that |x-y| \lt \delta \implies |f(x) - f(y) < \varepsilon or f(B_{\delta}(x)) \subset B_{\varepsilon}(f(x)), or B_{\delta}(x) \subset f^{-1}(B_{\varepsilon}(f(x))) = \{ y \in \mathbb{R} | f(y) \in B_{\varepsilon}(f(x)) \}.
Definition
The definition is defined at: Continuous.
Proposition
A Continuous f : X \to \mathbb{R} with X Compact, it will attain both a maximum and a minimum.
Proof
Note: the Continuous image of a Compact is Compact since Inverse Images of an Open Cover will again be an Open Cover. Then the final step is to use the Heine–Borel theorem since f(x) is compact in \mathbb{R}.
QED.
Something else
We say f is Continuous (at every x) if f^{-1}(A) is Open Sets for every open A \subset Y.
Continuous images of Connected spaces are Connected. Again because inverse images of Open Sets disconnecting an image would disconnect the domain.
So homeomorphisms preserve compactness and connectedness. They are Topological Invariants.