Fig. 123.
The triangles are shown in [fig. 123] as they join on to the triangle in the ochre cube. But they join on each to the other in an exactly similar manner; their edges are all identical two and two. They form a closed figure, a tetrahedron, enclosing a light brown portion which is the portion of the cutting space which lies inside the tesseract.
We cannot expect to see this light brown portion, any more than a plane being could expect to see the inside of a cube if an angle of it were pushed through his plane. All he can do is to come upon the boundaries of it in a different way to that in which he would if it passed straight through his plane.
Thus in this solid section; the whole interior lies perfectly open in the fourth dimension. Go round it as we may we are simply looking at the boundaries of the tesseract which penetrates through our solid sheet. If the tesseract were not to pass across so far, the triangle would be smaller; if it were to pass farther, we should have a different figure, the outlines of which can be determined in a similar manner.
The preceding method is open to the objection that it depends rather on our inferring what must be, than our seeing what is. Let us therefore consider our sectional space as consisting of a number of planes, each very close to the last, and observe what is to be found in each plane.
Fig. 124.
The corresponding method in the case of two dimensions is as follows:—The plane being can see that line of the sectional plane through null y, null wh, null r, which lies in the orange plane. Let him now suppose the cube and the section plane to pass half way through his plane. Replacing the red and yellow axes are lines parallel to them, sections of the pink and light yellow faces.
Where will the section plane cut these parallels to the red and yellow axes?