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Let forlet be an increasing function (i.e. if).Prove that if the sequence converges, to then isincreasing, and that if is continuous then the convergence is uniform.

  1. We first seek to prove that if the sequence converges to then is increasing.

Let such that. Since,thensuchthat ,




If we let then we have and.Implying that.Since we can make as small as we want, holds.

  1. Secondly, we have to show that if is continuous then the convergence is uniform.

We will prove by contradiction. That is, we assume that iscontinuous but the convergence is not uniform. This implies that such that where isa function of and where for.Due to being continuous, diverges to as.The fact that leads to a contradiction if we consider the value of suchthat and.Therefore isuniformly continuous.


The graph of is the set of

  1. If is connected and is continuous, prove that the graph of is connected.

We’ll have to show that the image ()of a connected set ()by a continuous function is connected.

We’ll proceed to prove the contrapositive. That is if ()is not connected then ()is not connected. If ()is not connected, then with, , (i.e. they are disjoint) and .

Considerand note that these two separateby definition of open sets. We know that since and similarly.Again since we have that.

Finally, sinceimplies implying that or.Thus is not connected.

  1. Give an example to show that the converse is false.

Let.Now but at.Therefore such that

  1. If is path-connected and is continuous show that the graph is path-connected.

Consider Then, since isconnected a continuous curve such that and.Then the curve is continuous and maps.

  1. What about the converse?

The converse is not necessarily true if we consider to be a discrete space with more than one point (for it to bedisconnected, it has to have more than one point) that maps to theconnected set.


  1. Prove that a continuous function, all of whose values are integers, is constant provided that is connected.

We will prove the contrapositive. Suppose such that (i.e. are distinct values of).Then we have andare both non empty disjoint open sets since all values come from adiscrete space of integers such that since andare arbitrary integers. This implies that if is not constant, then is not connected (the contrapositive).

  1. What if all the values are irrational?

Similarly, will have to be constant since is only connected if and only if is single valued. This is due to existence of the rational numbersthat allow to be written as a union of disjoint open sets when.