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Let (X,d) be a metric space, and let A be a subset of X_ We shall consider A with the subset metric dA- Assume that G = A is open in (X,d). Show that G is open in (...

Question

Let (X,d) be a metric space, and let A be a subset of X_ We shall consider A with the subset metric dA- Assume that G = A is open in (X,d). Show that G is open in (A,dA): Find an example which shows that although G € A is open in (A,dA) it need not be open in (X,dx). Show that if A is an open set in (X,dx). then a subset G of A is open in (A,dA_ if and only if it is open in (X dx)

Let (X,d) be a metric space, and let A be a subset of X_ We shall consider A with the subset metric dA- Assume that G = A is open in (X,d). Show that G is open in (A,dA): Find an example which shows that although G € A is open in (A,dA) it need not be open in (X,dx). Show that if A is an open set in (X,dx). then a subset G of A is open in (A,dA_ if and only if it is open in (X dx)



Answers

PROVE (Closure under Linear Combinations). If S is a finite set of vectors in a vector space V , then every linear combination of S is also a vector in V.

In the problem we have This integration 0 to a fx. Just relax. General S X dx. Now this is equal to what. So this is equal to let us consider this as you And this as which is. So this is of the form islet that is inverse logarithmic algebraic, programmatic and exponential. So let us assume you as the first function and we s the second Frandsen according to this. So we have this is effects immigration do double ds X dx minus integration. D upon dx off fx integration. General does X. Dicks hold X. So further this can be read in us Effects judas X minus integration. This is F. S. X. Into judas X. Dicks. Further we have to solve this spot. So it becomes F L. S X. Division G D S X dx minus integration. Day upon the excessive every day. Six integration Due to six DX holy X. So this is equal to every six and two jx minus integration if they were last X into gx dicks. So this is the different integration of the spot. So these every sex into JD six. The ecstatic was this one. So further we have the digit of overall integration. So this is overall integration will become now first of all we neglected so we neglect if true and G row as these are equals to zero. So what all integration become integration and you go to a fx G doubled us X dx at equals two F f. A judas A minus. So this is the stamp F or fe into Judith a minus. This is if does uh, g of a minus integration zero to A. If nevertheless X G of x dx. So this is the overall integration. Hence it is the answer to the problem.

Let's I be an open in trouble. So a function which belongs to this open and terrible will be an increasing function. If, for any X one viable X one extra which belongs to this open interval, I there exists F x one, which will be less than f off extra always So in that case, still say that the function F is increasing on the open in double I so therefore, the world that used to fill in the blank is increasing.

The graph of one by X F of x is equal to one, bikes will be and rectangular hyperbole something like this, sorry. So this is a graph of one banks now minus four to minus one minus four minus one means somewhere here minus four year minus one here. Obviously the graph is continuous and differential because it's very smooth from here to here. So it satisfies the conditions. The mean value system satisfies the conditions continuing and differential Pretty right now. So the function is run by ex we need to find the value of cyst that F f b minus F of a, divided by b minus a is equal to f dash of C. And what is your A or a or the left hand point minus four. B. Here is the right hand point minus one. So that means f minus four. I'm sorry f of minus one minus f. Of minus four, divided by b minus eight minus one minus of minus four is equal f dash of C. So what is the Director of One bikes? One bikes is basically expand minus one by powerful minus one. Expo minus one minus one. So it's minus one by its square. So that means minus one by c square. What is it? About minus one? It is the reciprocal of minus one minus one by minus four, divided by three is equal to minus one by c square, so minus one by c square is equal to uh one by four minus one by three. So minus one by c square is equal to minus three by four, divided by 31 by c square is equal to one by four. Cc square is equal to four. So C is equal to negative two because positive we can take because negative two is only one which is lying in between these two points. All right, that's it. That's the value.

This question covers topic relating to linear and zebra and the span of uh basis of uh vector space. So you can see here we have the set as a finite and had the route that the the linear combinations of all vectors of S. Is actually there's a set of fee. Okay, so how do we do that? First of all? You refine the uh you can say like we can pick a finite set um the subset of S. And that obviously ideas upset of the vector space V. And because it's the it's an element in sign as actually inside V. So that ai can be rewritten as B J E J. What do you Hj? So E J is um it can be finite or infinite is a basics the basis for V. Okay, so we um so as you can see here um a I can be rewritten as a linear combination of these spaces. Right? And now if we pick any vector, like for example, I pick victor, A and A. Is A and finally linear combination of ai right? So for example, and for I ai right from one, for example, from one to let's say end and maybe uh M. Okay. And so we can re written that some as and for nighttime submission of changing from one to N. B to J. J. And death is just a linear combinations. Uh The finite clinical combination of or the vector E. Uh J. Right? So if you want to write it down precisely, it's just the summation from I from one to M. And information from che from one to end on. For I B J E K. Right? And if you put out the so ehh Okay. So now if you put out each A out put E. J. As you can okay, you can interchange this summation to change this to end and you change it to em. And and for I. B. J. E. J. Right? Okay. So now just uh J equal to one to end of E. J. Um, submission I from 1 to 1 for I beta J. Okay, So this is just a real number, right? So it's a linear combination, so that means I belong to the span of ass. Or so that means the span of S. Is upset of fee.


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