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What differences are there between the following properties:

$(\forall j \in \{1,2,3\}) ( \exists a \in \Bbb R) f_j(a)=1$

$ (\exists a \in \Bbb R) (\forall j \in \{1,2,3\}) f_j(a)=1$

$(\exists j \in \{1,2,3\}) ( \forall a \in \Bbb R) f_j(a)=1$

$ (\forall a \in \Bbb R) (\exists j \in \{1,2,3\}) f_j(a)=1$

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SAM.Am
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  • It says that the function definitely takes the values $1$, $2$, and $3$. It does not tell you how to graph, because it does not tell you who is mapped to $1$, just that at least one element of $R$ is, etc. – Arturo Magidin Sep 04 '19 at 18:25
  • I update the question and would like to understand the difference between all of those properties. Thanks. – SAM.Am Sep 04 '19 at 18:40
  • @SAM.Am What do you think on your side of the meaning of those formulae? Have you tried to pickup functions that satisfy one but not the others? – mathcounterexamples.net Sep 04 '19 at 18:46

3 Answers3

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The first statement says: for each $j$ in $\{1,2,3\}$, you can find a real number $a$ (which number may depend on which $j$) with the property that the $j$th function maps $a$ to $1$.

Presumably, you have three functions called $f_1$, $f_2$, and $f_3$, and the first statement tells you that each and every one of these three functions "hits" the value $1$ at some point. It need not be the same point for every function.

The second statement says that there is a real number $a$ with the property that each of the three functions $f_1$, $f_2$, and $f_3$ send $a$ to $1$; here, $a$ is independent of the function: the same number $a$ has to "work" for all three functions.

The third statement says that there is at least one of the three functions that sends every real number to $1$; that is, at least one of the functions is the constant function $1$.

The fourth statement says that if you take a real number $a$, you will be able to find that at least one of the functions $f_1$, $f_2$, and $f_3$ send $a$ to $1$, and that this works for every real number $a$.

Graphically: imagine you draw the graphs of the three functions on the same plane. The first statement says that the horizontal line $y=1$ will intersect all three graphs, at least once for each graph; but it may intersect them at different points for each graph.

The second statement says that there is a number $a$ such that all three graphs go through $(a,1)$: the vertical line $x=a$ touches all three graphs at height $1$.

The third statement says that at least one of the graphs is the horizontal line $y=1$.

And the fourth statement says that every vertical line $x=a$ intersects at least one of the graphs at height $1$, but which graph it intersects may be different for different values of $a$.

Arturo Magidin
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  • Yes, this is exactly what our teacher did but I did not manage to grab the meaning of it. Well, I am in a French university and I studied math in English in an American school all my life (btw I did not get those concepts in my IB). So I am fighting with new concepts and in French too. Thanks A LOT. – SAM.Am Sep 04 '19 at 18:56
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    @SAM.Am: I'm confused; if I'm doing exactly what your teacher did but you did not get it then, are you having difficulty with this as well? (Is the final statement sarcasm?) Or is it that you were having language difficulties and couldn't quite follow but now that it is written out in English the penny has dropped? (For what it may be worth, you may have to re-learn math in the new language; I essentially had to re-learn calculus in English after learning it in Spanish to be able to reason about it in English, and there is group theory stuff I just cannot think of in Spanish...) – Arturo Magidin Sep 04 '19 at 18:59
  • In fact it's a long story, basically I came from a Froncophone country, but I did all my studies in American School abroad and I had to take whole year to re-study French again, Our teacher did the same demonstration but it was fast and furious scenario as I as struggling with my Chinees French to catch the strategy. When I read you reply I found it accurate and as if I went through the explanation in a slow replay mode. Again, Thanks a lot for your clear and accurate explanation it helped me a lot. My dad is a Math aficionado and he told me about this helpful portal. – SAM.Am Sep 04 '19 at 21:34
  • You wrote at the end of your reply this:

    "And the fourth statement says that every vertical line =".

    I did not get it as I think there is only a point (a,1) and not a vertical line. am I right?

     – SAM.Am Sep 05 '19 at 05:12
  • @SAM.Am: At every value $a$, at least one graph takes the value $1$. So if you draw the vertical line $x=a$, that line will intersect at least one of the graphs at the point $(a,1)$.The graph it intersects only includes the point $(a,1)$ in that vertical line. – Arturo Magidin Sep 05 '19 at 05:20
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The first one says that for every $j$ in the set $\{1,2,3\}$, there exists an $a$ in the set of real numbers dependent on $j$ such that $f_j(a)=1$.

The second, that there exists an $a$ in the set of real numbers such that for every $j$ in the $\{1,2,3\}$, no matter which one, we have $f_j(a)=1$.

Can you take it from here?

Hint:

The third is like the first, except with $\{1,2,3\}$ and $\Bbb R$ swapped. The same is true of the fourth and second.

Shaun
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So you have three functions, named by the indices $\{1,2,3\}$, which accept real values as arguments (aka. their domain is points on the real number line).

$(\forall j \in \{1,2,3\}) ( \exists a \in \Bbb R)~~f_j(a)=1$

Each index has at least one argument which makes its function equal one. These arguments need not be the same.

Each function will equals one at some point. They need not do so at the same point.

$ (\exists a \in \Bbb R) (\forall j \in \{1,2,3\})~~f_j(a)=1$

There is an argument which makes the function of every index equal one. Such an argument must do so for all three functions.

At some point, all three functions do equal one simultaneously.

$(\exists j \in \{1,2,3\}) ( \forall a \in \Bbb R)~~f_j(a)=1$

For at least one index, its function is constantly $1$.

At least one function equals one everywhere.

$ (\forall a \in \Bbb R) (\exists j \in \{1,2,3\})~~f_j(a)=1$

Each real number argument will make at least one index's function equal one (not necessarily the same function at different points, but at least one of them at any).

At every point at least one of the three functions will equal one.

Graham Kemp
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