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Such That=IFF?

E01

New member
Dec 8, 2013
9
This may seem like a dumb question but I'm not sure whether "such that" is equivalent to "iff" or "if then".

Here is what confused me. The image of A "f(A)" is defined as y element of Y such that for some x element of A, y=f(A). I could say there exists some element x such that y=f(A).

I'm not sure if I can say in a proof that if y=f(A) then there exists some x element of A.

Sometimes it seems like A such that B means "if then".
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Here is what confused me. The image of A "f(A)" is defined as y element of Y such that for some x element of A, y=f(A). I could say there exists some element x such that y=f(A).
This is not a correct definition. For one, you say "x such that" and then don't mention x. Could you give a correct definition and start by saying what $f$, $A$ and $Y$ are?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
This may seem like a dumb question but I'm not sure whether "such that" is equivalent to "iff" or "if then".

Here is what confused me. The image of A "f(A)" is defined as y element of Y such that for some x element of A, y=f(A). I could say there exists some element x such that y=f(A).

I'm not sure if I can say in a proof that if y=f(A) then there exists some x element of A.

Sometimes it seems like A such that B means "if then".
In my experience, "such that" usually occurs in THIS setting:

$S = \{x \in T \text{ such that } P(x)\}$

where $P$ is some property $x$ has, in other words:

$x \in S \iff (x \in T) \wedge P(x)$

For example:

$2\Bbb Z = \{k \in \Bbb Z \text{ such that } 2|k\}$

which says two things:

1. $k$ is an integer
2. $k$ is divisible by 2.

In this example, "such that" doesn't play the role of "if...then", "only if" OR "iff", it plays the role of "and".
 

E01

New member
Dec 8, 2013
9
Thanks for the response Deveno. That makes sense. So I would say y is an element of Y and y=f(A) for some x element of A.

In response to Evgeny.

I forgot the starting portion of the definition. Let X and Y be sets. Let f be a function from X onto Y. A is a subset of X. We define the image of A as the set f(A) where y is an element of Y such that y=f(A) for some element x element of A.

I need to learn to use LaTeX.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
In response to what I THINK you were trying to ask:

If f(A) = Y, then yes, there is SOME x in X with f(x) = y, for any y in Y you care to choose. This is often taken as the definition of "surjective" or "onto". Such an x is called a "pre-image" for y.

If, however, f(A) is not all of Y, then we cannot say this necessarily.

Example:

The function f:{1,2} --> {1,2} defined by:

f(1) = 1
f(2) = 1

is not onto, there is no pre-image of 2.

The function f:{1,2}-->{1} defined with the same values IS onto, as there is only one element in the co-domain, 1, and it has a pre-image (actually, it has two).

Any function f:X-->Y can be made into an onto function by considering f (with the same values at every x in X):

f:A --> f(A).

This underscores something that is often overlooked with functions: the "target-set" (the co-domain) is part of the definition, and f is not determined SOLELY by its values.

If all one is concerned about is f(x) for various x's, as in calculation of maximum or minimum values, often one does not think much about this, but it can be critical.