Analysis

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The definition of a connected set is a set which can NOT be partitioned into two disjoint open subsets and such that .

The Cantor set is famously disconnected, but how can this be? After all, the Cantor set consists of a bunch of isolated points, so how is it possible to even define an open subset of the cantor set without leaving the set?

Also, when proving a function is continuous, how do you know whether to use the standard definition or the "inverse function pulls back open sets to open sets or closed sets to closed sets" definition? Or can you use either of them as a matter of preference?

thanks

[kamil9876]

Also, when proving a function is continuous, how do you know whether to use the standard \epsilon-\delta definition or the "inverse function pulls back open sets to open sets or closed sets to closed sets" definition? Or can you use either of them as a matter of preference?

yep they are equivalent, as to which is simpler it depends on the information you are given.

The Cantor set is famously disconnected, but how can this be? After all, the Cantor set consists of a bunch of isolated points, so how is it possible to even define an open subset of the cantor set without leaving the set?

Any set of isolated points is in fact disconnected (it's even true in the english sense of the word). Just place some open interval around some subset of the points. An example for the Cantor set is: It's clearly open. It's complement is too. (Btw, perhaps you find this confusing because you're thinking about all those other real numbers, if you are treating the Cantor set as THE metric space you are working in then those other numbers don't really "exist" as far as you're concerned, the cantor set is a disconnected metric space(disconnected "set" sounds more ambigous I agree, since it doesn't specify what the space is)).

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Hmm thanks Kamil, I find the whole notion of 'open sets' quite confusing.

When people just refer to 'open sets', do they mean open in a metric (i.e. neighbourhood definition), or open in a metric subspace?

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If you know that a function is continuous, is there a way to use that to show that it's inverse is also continuous, without direct epsilon-delta on the inverse?

[humph]

If you know that a function is continuous, is there a way to use that to show that it's inverse is also continuous, without direct epsilon-delta on the inverse?

Not necessarily, because a function may be continuous but not even have an inverse function - e.g. take given by . Even if a function is continuous and bijective onto its image (so that an inverse function exists), its inverse function can still fail to exist - for example, take given by (why isn't its inverse continuous? Think about it...).
In fact, a continuous function with a continuous inverse is called a homeomorphism, and they're a pretty important type of function. I suggest you look it up. In this case, it is often more useful to think about continuity topologically (i.e. in terms of pullbacks of open/closed sets) than analytically (i.e. limits of sequences or epsilon-delta). It depends on the application though, of course.

[kamil9876]

For continous one to one functions from reals to reals there are some simple ways of showing the inverse is continous. One is to do - what is graphically intuitive - to use IVT and monotonicity: Simply suppose . To show is continous at just take consider and . Supposing is monotonically increasing (it's either strict increasing of decreasing to be one to one, a consequence of IVT), now for all we have .
So you can just set

edit: have to divide by two

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Going over a proof of Bessel's inequality, I'm confused about the following bit:
Let and let be a countable set of numbers

Suppose that (where the sum is taken over an arbitrary finite collection of s). How do we know that ?

[humph]

Write

We are given that

By definition,

This is lesser than or equal to

which is clearly lesser than or equal to by the definition of the sequence .

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Thanks humph!

With bounded linear operators, , and

In Shakarchi's proof that , I understand the direction .

But in proving , the proof method used is simply to show:

.

How does this imply that ?

[humph]

By definition, . So if we show that there exists such that , it must be true that by the definition of infimum. Thus if , then .

So really it shouldn't be , but rather .

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Oh, thanks, I understand it now