# Discrete mathematics, bijections between disjoint unions

1. Oct 24, 2012

### infk

Hi,
So I am trying to show the following:
$(A \cup B)\sqcup(A \cap B) \leftrightarrow A \sqcup B$

The proof that I am trying to understand starts with:
$A \leftrightarrow (A \backslash B) \sqcup (A\cup B) \qquad (1)$,
and
$A \cup B \leftrightarrow (A\backslash B)\sqcup B \qquad (2)$

These two both make sense, in the first one we see that $(A \backslash B)$ with $(A\cup B)$ is a partition of A, and thus there are bijections from $(A \backslash B)$ to $\{ (x,0) : x \in (A \backslash B) \}$ and from $A \cap B$ to $\{ (y,1) : y \in A \cap B \}$, and the same argument for $(2)$

The next step in the proof says that:
$(A \cup B)\sqcup(A \cap B) \leftrightarrow (A\backslash B)\sqcup B \sqcup (A \cap B)$ which (apparently) follows from $(2)$
and then, $(1)$ apparently means that
$(A\backslash B)\sqcup B \sqcup (A \cap B) \leftrightarrow A \sqcup B$ which proves it.
But I don't understand how $(1)$ and $(2)$ helps us in these steps, can you just exchange terms like that? Anyone care to explain? Thanks..

Last edited: Oct 24, 2012
2. Oct 25, 2012

### stauros

What the"$\sqcup$" stand for??

3. Oct 25, 2012

### infk

$A \sqcup B$ is the disjoint union of $A$ and $B$, defined by:
$A \sqcup B = \{(x,0):x \in A \} \cup \{(y,1):y \in B \}$

4. Oct 25, 2012

### Erland

And $\leftrightarrow$ ? Does that simlpy mean equality (=) here?

5. Oct 25, 2012

### infk

I thought this was standard notation in (discrete) mathematics, but anyway,
$A \leftrightarrow B$ means that there exists a bijection between $A$ and $B$.

6. Oct 26, 2012

### infk

So there is not a single person on the board who knows about this? It is from a first course in discrete mathematics for 2nd year students.

7. Oct 26, 2012

### ImaLooser

I can't get your example to work when using this definition. I don't get it at all.

8. Oct 26, 2012