Lecture 7 – Uniform Continuity

# Lecture 7 – Uniform Continuity

Now let us see some more applications of the
intermediate value property. It starts with the
following problem. The question is as follows . Does there exist a continuous function f from
R to R which has the following property, such that
if x is a rational number, that is, x is in Q, then f x is an irrational number. That
is, it belongs to R minus Q and if x is an irrational number,
that is, x belongs to R minus q, then f x is a rational
number, that is, f x belongs to Q. So is the question clear? We are asking for, does there
exist a continuous function f from R to R, such that,
if x is rational then f x is irrational and if x is
irrational, then f x is rational. So we proceed with the problem as follows: Let us define a function g from R to R by
the following rule. What is g? I define g x is equal to f
x plus x, for all x in R and we are trying to give the contra positive argument. That
is, assume 1 .that f is continuous. So I want to get a
contradiction out of it and then that will show no such a
continuous function exists. . If I assume that f is continuous, then the
first thing which follows is that g is continuous. Now in
the following arguments, I am going to use one thing time and again. That is this: if
x is a rational number and y is an irrational number,
then x plus y is an irrational number. This is the
observation which I am going to use time and again which is very easy to see because if
x plus y is in Q, that means it is a rational number.
Then x plus y minus x, that is also in Q. Let us write it
down. If x plus y belonging to Q, then y equals to x plus y minus x belonging to Q, which
is a contradiction simply because that sum and
product of two rational numbers is a rational number. This is the observation, if I use then I get
that sum of a rational number and an irrational number
is always an irrational number. This I am going to in my arguments. Now if I have g
is continuous, let us look at the property of
g. Suppose x belongs to Q. Then I know the defining
property of f that f x belongs to R minus Q but then this implies, if I call this star,
by star, g x, that is equal to f x plus x, that is an irrational
number. That means, it is in R minus Q. If x is
rational, g x is irrational. Now let us take x to be an irrational number. 2 .Suppose x is in R minus Q, then I know again
that f x belongs to Q but then this implies again by
star that g x, that is equal to f x plus x, again belongs to R minus Q. So it follows
from these two calculations that g has the following property
that whatever x you choose in the set of real number, g x is always an irrational number.
That means what I have is g is the map from R to R
minus Q and g is continuous. That I am getting because I have assumed that f is continuous. . Now let us see what more can I say, more about
it. Let us try to use the intermediate value property on the function g. Let us take some,
a strictly less than b, where a, b, both are real
numbers. Then let us look at g a and g b. It can happen that they are same and it can
also happen that they are not same. That means, let us
assume this. Now I know that g a and g b, both are
irrational numbers. Then what does intermediate value property say? Suppose I choose an x
which lies between g a and g b. So x lies between g a and g b and let us choose x to
be a rational number. That I can do because if g a and g
b are unequal real numbers, then always there exist a
rational number in between those two. By the intermediate value property as g is
continuous, by the intermediate value property, I write
it as I V P. There exist y in R, such that g of y is equal to x. But then you see, I
got a contradiction because g y is standing out
to be a rational because I have chosen my x to be a 3 .rational number but look at number the definition
of g. It is map from R to the set of irrational number R minus Q. That means, for no x in
R, g x can be a rational number. That has happened.
It has happened simply because I have assumed g a is not equal to be g b which is not true.
because g cannot take rational values, implies g a is equal to g b. What does this mean because I have a b to
be arbitrary real numbers? This implies then that g is
constant function. That means, that is g x is equal to a number c for all x in R and
surely c has to be an irrational and c is an irrational number.
Now, since I have got that g x is equal to c, I got
some inference about f also. (Refer Slide Time: 10: 07) Just writing down the definition of g I get
that g x is equal to f x plus x, that is equal to c for all x
in R and always remember that c is in R minus Q, that is, it is in the irrational number.
Then what do I know about f? That means f x is
just a line. That is, it is c minus x. Now in particular
let me choose x equals to c by 2. Then I know that this is also an irrational number. This
cannot be rational because if c by 2 is rational,
call it r. Then, c is equal to 2 r but if r is rational, then 2 r
is rational. That means, c will be rational which is not the case. That means c by 2 must
be an irrational number but what happens to f x? 4 .What is f of c by 2, that is c minus c by
2? That is equal to c by 2. Notice that c by 2 is an
irrational number and f of c by 2 is c by 2, which is also an irrational number then.
But you see that cannot happen. f takes rationals to irrationals
and irrationals to rationals, but in this case
what is happening? f is taking an irrational number c by 2 to the same number c by 2, which
is an irrational number which is a contradiction. This implies that f is taking an irrational
number, c by 2 to the same number c by 2. That is a
contradiction to the defining property of f; it says that f takes rationals to irrationals
and irrationals to rationals. So this is a contradiction
which implies, f cannot be continuous. That solves our problem. Now, let us see another
application of the intermediate value property. . Let P x stand for a polynomial. We know how
do polynomial looks like. It is a 0, plus a 1 x, plus
a 2 x squared, plus a n x to the power n and I say a n is not equal to 0 and assume a n
is positive. Now I say that if n is odd, that means it
is of the form 2 k plus 1. Then P has a real root. What
does that mean? That means there exist some x naught in R such that P of x naught is equal
to 0. What I want to do is, I want to apply intermediate
value theorem to the function P x because I
already know that any polynomial function is actually a continuous function. 5 .Fine. What I do first is, I write P x as
x to the power n a n plus a n minus 1 divided by x plus a 0
by x to the power n. The naive idea goes as follows that if I look at the quantity, if
you take x large, it does not matter in which direction,
positive large x or negative large x, you feel, that this
quantity should converge to a n because all the individual quantity which are involved,
they are having x in the denominator and that is getting
large. So those quantities will be small. So you
expect that the underlined portion converges to a n, which is positive. Now since n is odd, if I take x as positive,
then anyway x n is positive. If I take x as negative
then x to the power n is negative. That means, you expect, there will be a change of sign
for different values of x and then 0 comes in
between those different values. That means by the
intermediate value property there should exist x naught such that P of x naught is 0. That
is the idea we want to use. But we have to make it
rigorous. Let us first write down. What does it mean
to say the portion in the parenthesis converges to a naught? I say that given epsilon bigger
than 0, I choose it in such a way that a n minus epsilon
is positive. I say there exist a real number k in R, such
that, a n minus epsilon is less than a n plus a n minus
1 divided by x a 0 by x to the power n, which is less than a n plus epsilon for all x in
R such that, mod x is bigger than k. That means the k I
have chosen, it is a positive k. This is just meaning of
the fact that as x goes to infinity, the portion in the bracket converges to a n. I have just
written it in terms of epsilon following the definition
of limit. Now if I choose x bigger than k, what
happens then? Then, a n minus epsilon times x to the power
n is anyway less than x to the power a n plus a n
minus 1 by x plus a 0 by x to the power n, which is less than a n plus epsilon times
x to the power n. Notice then that by my choice this
also happens because a n minus epsilon is positive. I
have taken my x to be positive. That is also bigger than 0. Now what I can do is, suppose I choose x less
than minus k. That means, now I am choosing x to
be negative. Notice that since I have taken my n to be odd, x to the power n is also negative
if x is negative. So what does this imply? 6 .. This implies that a n minus epsilon times
x to the power n is bigger than x to the power n times a
n plus a n minus 1 by x plus a 0 by x to the power n, which is bigger than a n plus epsilon
times x to the power n. Now notice that, as x is less
than minus k, that is, x is negative, this implies that x
to the power n is strictly less than 0. This implies, now since x is less than minus k,
it implies x to the power n is also strictly less than
0. That implies what? It just implies that x
to the power n a n plus a n minus 1 by x plus a 0 by x to
the power n. This is less than x to the power times a n minus epsilon, which then is strictly
less than 0. So this implies what? This implies
that if x is bigger than k, then P x is bigger than 0. If x
less than minus k, then P x is less than 0. Now this implies then, there exist x naught
in R, such that, P x is equal to 0. Why so because there
is one value of x, for which P x is positive. There is
one value of x P x is negative. That means 0 lies between those two values and hence
by intermediate value property, there exist an
x naught such that P of x naught is equal to 0 and this
implies x naught is a real root, which is exactly what we wanted to prove. 7 .. Now we are going to start with concept called
uniform continuity. This is the concept which we
will interact while talking about Riemann integration of continuous functions. That
is the whole purpose of introducing this concept now. So
function and check the epsilon delta definition of continuity. First, what we do is we start
with the function example of function epsilon delta
continuity. So let us look this function, f x is equal
to 1 by x squared, where x belongs to open 0 infinity. I certainly cannot include the
0 in the definition because 1 by x squared will not
be defined then. What I want to check is first, this function
is continuous at every point of 0, infinity and to do
that what I will do is, I will just try to show that given epsilon, there exist a delta.
I want to see how do I find out delta and how does delta
depends on epsilon x, whatever. Choose a point x
equals to x naught to check continuity at x naught. How are you going to do about this
problem? I first look the quantity f x minus f x naught
and then just I write down the definition of f. So that
is modulus of 1 by x squared minus 1 x naught squared and then if I expand it what I get
is, mod of x minus x naught times x plus x naught.
I am not putting modulus here. Both x and x naught
are positive divided by x squared times x naught squared. 8 .Now let us assume that x satisfies the inequality
x minus x naught is less than x naught by 2.
That is, x belongs to x naught divided by 2 and then thrice x naught divided by 2. That
is a choice of neighborhood but if I have this,
then what I have is that mod of x that is equal to mod
of x minus x naught plus x naught and then I use triangle inequality to get x minus x
naught plus x naught but x minus x naught according to
my choice is less than x naught by 2. So this is less
than x naught by 2 plus x naught, which is equals to thrice x naught by 2 and then this
implies that mod of x plus x naught, by triangle inequality
is less or equals to mod x plus x naught mod x
is less than 3 x naught. So this is less than thrice x naught by 2 plus x naught. That is,
5 x naught by 2. . Now, if I start with mod x naught, which I
can write as mod of x naught minus x plus x and
apply triangle inequality, that is x naught minus x plus x, which then strictly less than,
what was the estimate on x naught plus x? Let us see.
x naught plus x is 5 x naught by 2. What I do is, I
look at x naught minus x plus mod x. But mod of x naught minus x is less than x naught
by 2. So what I have is here, I have mod x naught by
2 plus mod x. Actually many of those mods are
unnecessary because x and x not are positive. 9 .Anyway, this implies then that mod x is bigger
than x naught minus x naught by 2, which is
equal to x naught by 2. Now I want to use all these information in my estimate. That
is, mod of f x minus f x naught, which I know is mod x
minus x naught into x plus x naught divided by x
squared times x naught squared. Now I am going to use the above estimate. What I get first
is, this is less than mod x minus x naught. I
am going to do anything with it. We just keep it in the
rough form and I know x plus x naught by previous work that is less than 5 x naught by 2 and
in the denominator, what I am going to use is
that mod x is bigger than x naught by 2. That means,
1 by x squared is less than x naught squared by 4. What I am going to use here is, x naught squared
by 4 and there is another x naught squared remaining here. What I get then is, this is
mod x minus x naught, which is intact times 10 divided
by x naught cube. All now I want is, given epsilon we want mod of x minus x naught times
n by x naught cube to be less than epsilon. If
I want to have this, this implies we should have
mod of x minus x naught is less than epsilon into x
naught cube divided by 10. Now what is the delta I am
going to choose? Certainly, the choice is obvious. I choose delta equals to the minimum
of x naught by 2 and the quantity epsilon x naught
cube divided by 10. Now the point to note here is
that the delta which I got, whose precise expression I have now here, it involves x
naught. At the same time, it involves delta. That means, given epsilon, when we want to
choose delta, it depends on epsilon and the point x
naught where you are checking the continuity of the function. Now notice one more thing
that when I get my x naught to be small, that implies
this minimum, which is appearing that will also
be small. 10 .. Let us note it down. From the
previous expression, it follows that for fixed epsilon but small x
naught, a smaller delta will work. This is what I got analytically, but I want to see
it from the picture if that is possible. Let us try to
draw the picture. This is the y axis. This is x axis. I have 0
here. Let us try to draw the curve for f x is equal to 1 by x squared. It looks something
like this. Now first, let us choose the point x naught
here. This is my x naught and I take fixed epsilon.
Here, I have f x naught. Here is where is my f x naught. Now notice, if I choose epsilon band around
f x naught, this is f x naught. An epsilon band, how
does that look like? Let us say it is here, it is here. That means, I want a choice of
delta such that f x lies in this band. Then the obvious choice
is, you put this down. It seems this delta, this kind
of neighborhood of x naught will work, for which I will delta get. Now let us draw another
copy of this picture. This is y-axis and this is
x-axis. The graph of the function looks like this. Let me
choose x naught here. That means I have chosen a smaller x naught. Then the corresponding,
this is my f x naught. It is much higher. That
is obvious because if x naught is small, f x naught has to
be big because it is 1 by x naught squared. Now I choose an epsilon around, which is the
same epsilon which I have worked. So in the picture, it will look something like this
and this, the same epsilon band, then what should work as 11 .delta? That means you just have to project
it, down below, that is somewhere here and here, you
will get somewhere here. That means, this is now your neighborhood. See the length of
neighborhood? This much here, the length of the neighborhood was this much. It follows
from the picture, if my drawing is perfect, that
for this smaller x naught, the delta is getting small. That
means delta does depend on x naught. That might be because of I got delta because
of my sloppy arguments. Analytically I need to
prove that there is not a single delta which will work for all x not bigger than 0 for
this function f x. So in the previous example it seems that
if I give you a epsilon, then the delta you are going to
get, it depends on epsilon as well as the point x naught. Uniform continuity means this
does not happen. That means, delta can be chosen independent
of x. The precise definition is as follows. . Suppose if I have a f from the domain of f.
That is a sub set of the real line to R. Then f is called
uniformly continuous on the domain of f. If given epsilon bigger than 0, there exist a
delta bigger than 0, such that whenever mod of x minus
y is less delta for x, y in domain f then mod of f x
minus f y is less than epsilon. That is, you take any two points x and y such that mod
of x minus y is less than delta where x and y are in
the domain of f then mod of f x minus f y is always less
than epsilon. 12 .So the thing to notice, first of all, that
uniformly continuity deals with domain of f. It is not about
points; it is about the set. This is number 1. Number 2 is an easy exercise which you
can easily check. If f from domain of f to R is uniformly
continuous, then f is continuous on domain of f.
Now what I want to show you is that the function which I started with, f x is equal to 1 by
x squared defined on 0 infinity is not really
a uniform continuity function. That means there exist
an epsilon for which no delta works. Let us try to see that next, so the example. . Later we will see an easy proof of this. I
look at the function f x equals to 1 by x squared, where
x belongs to 0 infinity. I say given epsilon is equal to 1, let us choose a delta naught,
which works for some x naught. Now I am going to
choose an x 1 and I would show this same delta naught does not work for x 1. So what I do
is, I choose x 1 is equal to delta naught by 2 and I
choose x is equal to delta naught, then modulus of x 1 minus x, that is, delta naught by 2
which is strictly less than delta naught and then modulus
of f x 1 minus f x. That is, modulus of 4 by delta
naught squared minus 1 by delta naught squared, which is 3 by delta naught squared. That is
bigger than 3, as delta naught is bigger than 1. This implies, mod of f x minus f x 1 is
not less than epsilon. This implies the function f
x equals to 1 by x squared is not uniformly continuous
on 0 infinity but let us do some more experiment with this. 13 .. Again, I look at the function f x is equal
to1 by x squared. Now let me assume that x is bigger
than or equal to a, which is bigger than 0. Now I can show this function is uniformly
continuous. What does it show then? It actually shows
that the concept of uniform continuity depends on
which domain of f you are working with. Let us see why is this uniformly continuous.
So I look at f x minus f y. That is, mod x minus y
mod x plus y divided by x squared into y squared, fine. That means, I have to find a delta which
works for all x and y and a given epsilon. Since x and y both are bigger than a, what
I get is, this is less than equal to mod of x minus y into
mod of x plus y divided by a to the power 4. If I can,
somehow show that mod of x plus y divided by x squared y squared is bounded on a infinity
by a constant M, then what I get is that mod of
f x minus f y is less or equal to M times mod x minus
y. If I can show that, then this implies that
given epsilon bigger than 0 choose delta to be equal to
epsilon by M. Now notice that this delta is not depending x and y because if I choose
my delta to be less than epsilon by M, this would imply
mod x minus y less than epsilon by M implies mod
of f x minus f y is less than or equal to M times epsilon by M, which is epsilon. So
all I need to 14 .show that this quantity is bounded by some
constant M which is independent of x and y. That is
pretty easy. . I just write mod x plus y divided by x squared
y squared as the mods are unnecessary because I
am working with positive x and y. Anyway, then this is 1 by x y squared plus 1 x squared
y, but as x bigger than or equal to a and y is bigger
than or equal to a. What I have is, 1 by x y squared
plus 1 x squared y is lesser equal to 1 by a cube plus 1 by a cube which is equals to
2 by a cube which I can take to be equal to my M. So this
implies mod of f x minus f y is less or equals to 2
by a cube times mod of x minus y. That means my choice of delta actually is, a cube epsilon
divided by M. See, delta depends on epsilon, not on the
points, but certainly depends on the set. Notice that if I
had taken a to be equal to 0, that was 0 infinity I wanted to work with. Then my delta will
be 0, so close to 0. I have got a problem, but away
from 0. If the function is defined for x bigger than
or equals to a, then it quickly become uniformly continuous. 15 .. Now I come to most important theorem of uniform
continuity and why we wanted do it. We have seen that if I have uniformly continuous function,
then function is always continuous. We have also seen, there exist continuous functions
which are not uniformly continuous. For example, f x
is equal to 1 by x squared on 0 infinity. That is a continuous function certainly I
have checked but it is not a uniformly continuous. But
there are certain domains on which continuity always
gives me uniform continuity. That is what I am going to prove now. That is the theorem. Let f from closed interval a b to R be a continuous
function. Then f is uniformly continuous. That
is, the statement of the theorem says that if I have a continuous function defined on
a closed and bounded interval, then the continuous function
actually is uniformly continuous. Let us try to
prove this. The proof actually uses the contra-positive arguments. Let us assume that f is not
uniformly continuous. Assume that f is not uniformly continuous. What does this mean?
Some function is not uniformly continuous. It means there exists an epsilon for which
no delta works. The definition of uniformly continuous
says that given any epsilon there is a delta satisfying some inequalities. It is not uniformly
continuous then it means that there exists an
epsilon for which there is no good delta. Let us just try to write it down. It implies
there exist epsilon bigger than 0 such that, given any
delta bigger than 0, there exist two points which I will 16 .call x delta and y delta in the closed interval
a b, such that mod of x delta minus y delta is less
than delta but modulus f of x delta minus f of y delta is bigger than or equal to epsilon. That is what not uniformly continuous means.
Fine. With respect to this bad epsilon, I am going
to choose delta according to my will. So I will start choosing delta is equal to 1 by
n. Then corresponding to this delta there are bad
two points x delta and y delta which now I will call x n
and y n. Then there exist x n in a b and y n in a b, such that, mod of x n minus y n
is less than 1 by n but mod of f x n minus f y n is bigger
than or equal to epsilon. So the thing is, I have got hold of two sequences.
One is x n, other is y n on the closed interval a
b but I do not know anything about convergence of those sequences. But see, I am on the closed
interval a b and I know Bolzano Weierstrass theorem. It says that the given any sequence
in the closed interval a b, I have got a convergence
sub sequence. Let us consider convergent subsequence of x n, call it as x n k. . By Bolzano Weierstrass theorem, there exists
a convergent sub-sequence x n k of x n and let x n
k converges to x naught which is also in the closed interval a b
but since f is given to be continuous, I know that this implies f of
x n k converges to f of x naught. Also notice that mod of 17 .x n k minus y n k, that is less than 1 by
n k. What does this imply? It implies, x n k minus 1 by n
k is less than y n k which is less than x n k plus 1 by n k. Apply Sandwich theorem
now. This converges to x naught. This converges to x
naught. This implies, y n k also converges to x naught
but again, by continuity, this implies that f of y n k
converges to f of x naught. Then this implies that f of x n k minus f of y n k, that converges
to f x naught minus f x naught which is equal to
0. But if you look back the condition on f x n and f y
n was this: that is bigger than or equal to epsilon for all n but in that case it cannot
happen that it has got a subsequence which converges to 0
but that is what I have got here which is a
contradiction. This happens because I assume that f is not uniformly continuous. So it
follows that f is uniformly continuous. . Now let us ask the following question. Suppose
f from domain of f to R is a continuous function and let x n be a Cauchy sequence, then the
question I want to ask is, look at the new sequence f
of x n. The question is, is it Cauchy? That is, the question we are asking is that suppose,
I have given a continuous function, then is it true
if takes a Cauchy sequence to a Cauchy sequence? The
answer is no. Well, I look at this function f x equals to 1 by x squared where x belongs
to 0 18 .infinity. I look at this sequence, x n is
equal to 1 by n. Then this sequence is contained in the
domain of f certainly and also it is a Cauchy sequence. But then what is f of x n? f of x n is standing
out to be n squared which is an unbounded sequence. If you remember, we have proved
that the first thing a Cauchy sequence does is, it
becomes a bounded sequence. This is an unbounded sequence. It cannot be Cauchy. So f of x n
is not a Cauchy sequence. So in general, it is
not true that a continuous function takes a Cauchy
sequence to a Cauchy sequence but I am going to show if my function uniformly continuous,
then it does take a Cauchy sequence to a Cauchy sequence. Let us try to prove that. This is
the theorem. . Let f from domain of f to R be uniformly continuous.
If x n is a Cauchy sequence then the new sequence f x n is also a Cauchy sequence.
The proof is very simple. If I have to show f x n is a
Cauchy sequence, what I have to do is, to prove that given epsilon bigger than 0 there
exist capital N such that for all m, n bigger than
or equal to capital N mod of f x m minus f of x n is
less than epsilon. Now using uniform continuity of f, can get a delta corresponding to the
given epsilon such that if mod of x minus y is less
delta, then mod of f x minus f y is less epsilon. That
is the definition of uniform continuity but notice that x n is Cauchy. 19 .So thus there exists capital M such that
mod of x m minus x n is less than delta if m, n bigger
than or equals to M by the fact that x n is Cauchy. But then by definition of uniform
continuity mod of f x m minus f x n is less than epsilon
for all m, n bigger than or equal to the same capital
M. But this is precisely what I wanted to prove here, that given epsilon bigger than
0, there exist M such that, for all m, n mod of f x m minus
f x n is less than epsilon. See what I have found out.
I found out that mod of f x m minus f x n is less than epsilon for all m, n bigger than
or equal to M. That means, the stage N we called in the
previous case, I found it to be M and what is this M?
This M works for the delta in the Cauchy sequence x n. This proves the result. Now using this, it follows quite easily that
the function f x is equal to 1 by x squared is not
uniformly continuous in 0 infinity. Let us see how so. . I look at the function f x is equal to 1 by
x squared. Let me take x in 0 1. There also it is
uniformly continuous. If f is uniformly continuous, then x n Cauchy implies f x n is Cauchy.
That is not the case. Let me look at the sequence x n is equal to 1 by n which I have looked
at earlier. If x n is equal to 1 by n then x
n is a Cauchy but f of x n equals to n squared which is not 20 .Cauchy. But my previous theorem says if f
is uniformly continuous then a Cauchy sequence will
be mapped to a Cauchy sequence by f. I have assumed my f to be uniformly continuous
on 0 1 and I can see there exist a sequence 1 by
n which is not getting map to Cauchy sequence. That means f cannot be uniformly continuous.
So now then we have an easier proof of the fact that the function f x is equal to 1 by
x squared is not uniformly continuous in 0 infinity. Earlier
we have established the same fact using the epsilon delta definition of uniform continuity
but here we have an alternative proof using definition of uniform continuity. 21 .

## 10 Replies to “Lecture 7 – Uniform Continuity”

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8. Anil Kumar says:

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9. Anil Kumar says:

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