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March 1

Value of delta

Find a value of delta in terms of epsilon such that $|x-2|<\delta \Rightarrow \left|{\frac {1}{x}}-{\frac {1}{2}}\right|<\epsilon$

$\left|{\frac {1}{x}}-{\frac {1}{2}}\right|$

$=\left|{\frac {2-x}{2x}}\right|$

$\leq \left|{\frac {\delta }{2x}}\right|<\epsilon$

but there is still a pesky x in the denominator. Widener (talk) 11:36, 1 March 2011 (UTC)[reply]

Hint:

\delta =\min(1,2\epsilon )

works. First prove that this is indeed the case; then figure out how you could have found it without knowing it in advance. -- Meni Rosenfeld (talk) 12:13, 1 March 2011 (UTC)[reply]

The answer can be easily seen on a graph, but if you like to solve your problem by algebraic means I'd suggest to plug a new symbol into your expressions. Define $y=x-2$ which implies $x=y+2$ and ${\frac {1}{x}}-{\frac {1}{2}}={-y \over 2(y+2)}$ . Then the question reads: 'Which y satisfy the inequality $\left|{-y \over 2(y+2)}\right|<\epsilon$ ? What is their absolute values' upper bound δ in terms of ε?'
That can be transformed into a pair of linear inequalities (|m| < q ⇔ −q < m < q) — but don't forget to consider intervals, where denominator is positive and where it's negative, because 'less than' becomes 'greater than' on multiplying both sides by negative value. --CiaPan (talk) 06:41, 2 March 2011 (UTC)[reply]

Self-homeomorphisms and fundamental groups

Let X be a topological space. When $f:X\to X$ is a homeomorphism, then $f_{*}$ , the induced homomorphism on the fundamental group, is an automorphism. Is the converse true? I.e. if $f_{*}$ is an automorphism, then is f homotopic to a homeomorphism? I know it's true for some nice spaces (tori). Is this true for all spaces? For compact manifolds? Any reference discussing this would be appreciated. Staecker (talk) 17:27, 1 March 2011 (UTC)[reply]

No. Something homotopic to a homeomorphism has to induce an isomorphism on homology, cohomology, higher homotopy and so on, while inducing an isomorphism on the fundamental group is a pretty weak condition, especially if the fundamental group is trivial. For example, any self-map of S² with degree not 1 or -1 (such as a constant map, say) is a counterexample. Algebraist 18:15, 1 March 2011 (UTC)[reply]

Yes I should have thought of that- thanks! Staecker (talk) 18:34, 1 March 2011 (UTC)[reply]

Also, the fundamental group, and the other invariants Algebraist mentioned, are "homotopy invariant" (see homotopy), so any homotopy equivalence will induce an isomorphism on these. And being a homotopy equivalence is a lot weaker than being a homeomorphism. Aenar (talk) 18:57, 1 March 2011 (UTC)[reply]

But how does being a homotopy equivalence from X to X compare with being homotopic to a homeomorphism? Algebraist 19:01, 1 March 2011 (UTC)[reply]

Yeah, sorry, I just realized I hadn't read the question right, and that it is actually a good (hard?) question (especially if reformulated as "for which spaces is any homotopy automorphism homotopic to a homeomorphism"). Aenar (talk) 19:06, 1 March 2011 (UTC)[reply]

All I've discovered so far is that it seems to be a well-studied question. For example, the Borel conjecture is that the two are the same for aspherical closed manifolds. Whitehead's theorem already implies that any self-map of such a manifold that induces an isomorphism on the fundamental group is a homotopy equivalence, so that (conjecturally) gives a larger class of nice spaces than just tori. Algebraist 19:15, 1 March 2011 (UTC)[reply]

Interesting. Aenar (talk) 19:28, 1 March 2011 (UTC)[reply]

Yes very interesting- I hadn't heard of the Borel conjecture. Thanks a lot- Staecker (talk) 00:00, 2 March 2011 (UTC)[reply]